Found a solution! I'm not at all sure that it's optimal, meaning the trolls will eat me anyway, those assholes. In particular I don't make any use of the fact that there are three languages spoken by three trolls each [but stay tuned on that one, I've found a promising line of inquiry].
The number of questions is very variable depending on which answers the trolls give; I've counted
42 40 and it's likely to be a maximum (i.e. the actual number because the trolls would presumably time-travel to ensure it) but I didn't count the number of questions under all scenarios so it might not be.
(oh wait 42... Yeah it's the correct answer why did I even spend five hours on this)
[Later ETA: I found a way to shave one question off that total, but do I want the answer to be 41 or 42? I mean, how much do I NOT want to be eaten, really?]
BASIC TROLL TAXONOMY
We see the trolls as falling into two groups (ignoring the King Troll whom I will ignore entirely from now on): Trolls 2, 3, 6, 7 and 10 give answers that have a relationship to the question asked, i.e. they lie or tell the truth, while Trolls 4, 5, 8, 9 and 11 are all capable of ignoring the question for their answer. Both groups can further be divided: the "truthy" trolls will either behave as Knights, or as Knaves (10 is basically one of the two we just don't know which, while 6 could behave as neither but that would reveal them too easily so they will behave as one too. 7 I'm ignoring for a reason that will later become clear), while the "self-absorbed" trolls are capable of varying their answers (4, 5) or not (8, 9, 11).
A FEW WORDS ON RANDOM AND TELEPATHIC
In those four groups, the Truthy and Invariant trolls will reliably answer like they belong in their group, but the Variable trolls could answer anything so they are basically spoilers.
Moreover, a clarification from Victor Stafusa establishes that the Random Troll's random stream is "under the trolls' control" in the sense that it's one of the things they can go back in time to change to make whichever strategy I choose fail if possible, meaning the stream is not random at all, it's optimized to make this problem as hard to solve as possible.
The same can be said of the Telepathic Troll, because it can choose to answer whatever it wants anyway (as long as there are non-retired trolls that can answer "yes" and "no" at any point, which there are because 8 and 9 will never retire), or alternatively it can copy the Random Troll and its optimized-to-mess-me-up stream.
Either way 4 and 5 can be considered interchangeable.
ON SLICING CAKES
To divide a group into four I theoretically need only two questions (asked of each troll of course); however here I will not understand the answers they give, and because they do not all speak the same language I cannot use one's answer to deduce another's. But each troll does stick to its own language, so I can compare their successive answers to each other, which is what I'll do. This means however I need three questions, because two questions only yields one comparison.
These three questions will be:
To divide the Truthy group from the Invariant group:
Will your next answer be a lie? will be answered by "No" by all truthy trolls (2 tells the truth that it's "no", 3 lies that it's "yes" and so answers "no", 10 behaves as one of those and 6 has no choice but to give the only answer available to them), and as a bonus will make 7 go away because Undecided Troll cannot know whether it will lie or not next question, meaning it cannot form an answer to lie or tell the truth about.
Will your next answer be the truth?" similarly will be answered "Yes" by all truthy trolls; the Invariant trolls for their part will give the same answer to both questions. This means comparing the answers to both questions will separate out the Truthy from the Invariant;
And to tell the Knight-behaving truthy trolls from the Knave-behaving ones: Is 2+2=4? will do nicely. Trolls who give the same answer as to the first question are Knave-behaving (and, once we identify them as the Knave Troll or Meteorological Troll, we will know that that answer means "no") and Trolls who give the same answer as to the second question are Knight-behaving.
THE QUESTIONS COMMENCE
I actually start with the last question I mentioned, because that is a question that every troll can answer except the Silent Troll, meaning it allows me to identify him. [LATER ETA: I've just realized this isn't necessary: I can ask the questions in the order I first presented them, remember which two trolls retired from the first question and ask the Knight Troll later which is which, with 1 question. This allows me not to ask the last question to anyone who gave the same answer to the first two, saving me 2 questions (it won't be more for asshole-time-traveling-troll reasons), for a net gain of 1 question and a total number of 41, which is just plain silly honestly we all know the answer is 42)]. I will write all the answers the trolls give as a letter of the alphabet to represent that we don't understand their answer (and a different letter each since I don't know which speak which languages). So I ask each troll: "I): Is 2+2=4?" and I get the answers:
2 - A (Yes)
3 - D (No)
4 - E/F (Yes or No)
5 - G/H (Yes or No)
6 - I/J (Yes or No)
7 - K/L (Yes or No)
8 - M (Yes)
9 - P (No)
10 - Q/R (Yes or No)
11 - Goes away.
We've now identified Silent Troll and asked 10 questions (note that the Telepathic Troll wouldn't go away at this point because the Silent Troll can't talk but that doesn't mean they can't communicate telepathically).
The next question is : "II): Will your next answer be a lie?". The trolls reply:
2 - B (No)
3 - D (No)
4 - E/F (Yes or No)
5 - G/H (Yes or No) or goes away but really doesn't
6 - J (No)
7 - Goes away
8 - M (Yes)
9 - P (No)
10 - R (No).
So at this point either one or two trolls go away; if one does then I know that was the Undecided Troll. If two do then I know that the first one was the Telepathic Troll (channelling the Undecided one, which it can only do if Undecided Troll hasn't retired) and the second was the Undecided Troll. But this makes my life easier meaning the Telepathic Troll will do no such thing, so I've identified only Undecided Troll and asked 19 questions.
Finally I ask: "III): Will your next answer be the truth?" to the remaining eight trolls, for 27 questions. Their answers:
2 - A (Yes)
3 - C (Yes)
4 - E/F (Yes or No)
5 - G/H (Yes or No)
6 - I (Yes)
8 - M (Yes)
9 - P (No)
10 - Q (Yes).
Now I can divide the trolls into four (actually three) groups: the "Invariant group" are those that gave the same answer to all questions and contains 8 and 9 (and maybe 4 and 5, which could be in any group); those that gave the same answer to questions I and II but a different answer to III are "the Liar group", containing 3, 10 if the Meteorological Troll is a liar this week, 6 if it was channelling 3 for question I, and maybe 4 or 5; those that gave the same answer to I and III and a different one to II are "the Truthful group", containing 2, 10 if Meteorological Troll is Knightish, 6 if it was channelling 2 on question I and maybe 4 and/or 5. The last group, call it the "WTF are we even doing group", is those that gave the same answer to the "are you truthy" questions (so they're self-absorbed) but a different answer to the "which truthy are you" question (so they're not invariant); has no guaranteed members but could include 4 and/or 5. Call the groups I, L, T and W.
At this point there are a few things we can say. If L and T contain one or five members then we can use the guaranteed Knight or Knave Troll, and the fact we can guess "Yes" and "No" in their language from their previous answers, to identify everyone else [note this would take 11 questions at worst; this is relevant later]. If WTF group contains two members we can separate out the Knight, Knave, Joker and Meteorological Trolls using the question "Are you a Knight" (for members of the Truthful group) or "Are you a Knave" (in the L group), to which the Knight, Knave and Joker Trolls will give the same answer and the Meteorological Troll a different one. So, from the principle of maximum hardness on the part of the trolls we can assume the L and T groups contain three members each or two and four members. This implies that WTF group is empty and the Invariant group contains only 8 and 9.
SO WHAT DO WE DO NOW
So far, 2 (and all other members of T by definition) will have answered in the pattern "same-diff-same", and 3 (and all other members of L) will have answered in the pattern "same-same-diff". Call those patterns "SDS" and "SSD". Now if I ask all members of group T, "Are you the Knight Troll?" I am guaranteed that 2 (and 6 if it's there) will answer "yes", while 10 will answer "no", meaning if 10 is in group T I am guaranteed to end up with two patterns: SDSS and SDSD. I can do the same thing in group L with the question "Are you the Knave Troll?", yielding the patterns SSDS and (possibly, guaranteed only if 10 is in L) SSDD. Let's call the patterns that the Knight or Knave produces K-type patterns, and patterns the Meterologist produces M-type patterns. (I have now asked 33 questions).
IT GETS INTERESTING
At this point a few configurations can give it all away for me. If any group contains only one K-type pattern, then I've identified a Knight or a Knave and I know I'll get all the names eventually (forget optimal number of questions for now...). If there is only one M-type pattern, I've identified my Meteorological Troll and I know the weather on his planet last week so we're good. This leaves only a few configurations where I would lack information, i.e. a few configurations that are possible given trolls go back in time etc. If groups T and L have 3 members each, both groups will have two K-type trolls and one M-type troll. If the groups have 4 and 2 members, the 2-member group will have two K-type trolls, and the 4-member group will have two K-type trolls and two M-type trolls. Or put another way, we have 4 K-type trolls and 2 M-type trolls, with 2 K-type trolls in each group and the M-types trolls either both in one group or one in each group.
Now, if we have a 3/3 distribution, if I ask a troll in L or T group "Is the Meteorological Troll in your group?" (or, if the trolls don't like groups, ask "Is the Meteorological Troll one of this troll, that troll or that other troll?", it works out), the Knight, Knave, Joker and Meteorologist will all say the same thing, namely "yes" if 10 is in T and "no" if 10 is in L. The Joker is K-type so three out of four K-types will necessarily give the same answer, meaning that there will be one group where both K-types maintain the same pattern. That pattern will tell us whether the answer was "Yes" or "No" (for example, if it's in T-group and the pattern is SDSSS the answer is "Yes", because "S" is "Yes" for the Knight Troll), meaning we have unambiguously identified our Meteorological Troll, and we know whether it's lying or telling the truth this week, meaning we can identify all the other trolls.
Now notice that to come to this conclusion we only needed to interrogate the K-type trolls. This means that in the 4/2 configuration we can get the same result just by moving one M-type troll into the other group: if it happened to be the Meteorological Troll it would no longer give the same answer as 2, 3, and 6 because we'd have a truthful 10 in with the liars or vice-versa, and we need to take that into account once we know which group 10 is in, but since we only need to ask the K-type trolls it's irrelevant to finding where 10 is in the first place. In fact... we don't even need to ask all the K-type trolls! If we ask a group and both K-types give the same pattern then we're good. If we unluckily start asking the group in which one of the K-types is 4 or 5, this troll will break the pattern, meaning we won't know which pattern the actual 2 or 3 gave and will need to ask the other group. But since we know that in that group both K-types will follow the same pattern we only need to ask one. So, we have identified Meteorological Troll with 3 questions putting us at 36.
WRAPPING IT UP
Now all we need to do is ask the Meteorogical Troll questions like "Is this the Knight Troll?" and go from its answers, which we know are true or false depending on the group it's in and we know what's yes and what's no depending on its pattern of answers so far. We need 2 questions to identify the Knight or Knave in the two groups of two K-types; 1 question to identify the remaining M-type which is either Random Troll or Telepathic Troll; 1 question to distinguish Joker Troll from whichever of the two was the other K-type; and 1 question to identify the Positive and the Bad Humor Trolls (which have been waiting patiently in their group all this time, Invariant to the end; I hope Positive Troll found ways to help Bad Humor Troll to pass the time).
And so we get to 42, which is so obviously the correct answer I say we can stop here. It breaks my heart but now that I've shaved two questions from the total I need to face reality that Victor Stafusa did not, in fact, come up with the Ultimate Question to Life, the Universe and Everything :( Total questions so far is 40 (supposing we did kick out two trolls on the first question as described in the [later ETA] brackets). Theoretically we should count all the answers at the previous points of "now we have identified this troll we can trivially get the others" to make sure one of those paths doesn't force us to ask more questions than that, meaning the trolls might push us there. I've done that with the "there is only one troll in L or T" scenario, and that gets us to a mere 27+11=38 questions. I'll assume all other paths similarly need fewer questions.
NOTES AND APPENDICES
HEY, COULDN'T YOU KICK MORE PEOPLE OUT?
I don't see the point. 4, 8 and 9 will never leave, and 5 can never be forced to leave. 5 and 4 are the big spoilers, and the heart of my strategy is to identify a reliable truth-teller or liar so it can tell me what the others are; this is useless if I make the truthy trolls leave. I tried to see if weeding out the larger of the L and T groups could help, but there are two random spoiler trolls, insofar as there is one in each group (and therefore there will be one in each group) this strategy doesn't leave me further informed.
It should also be noted that it might look like asking questions that might make a troll go away provides more information than those that don't, because we have three potential answers ("yes", "no", leave) instead of two. However this is ignoring the kind of trolls we're talking about here. The answers "yes" and "no" as such provide next to no information from 4 or 5, and neither of those can be made to leave anyway. 6 cannot be made to leave in a way that won't also make 2 and 3 leave. "Leave/don't leave" can only be used to distinguish (2, 3, 6, 10) from (4, 5), and not even 5 actually because 5 can leave if it makes our lives harder. It is hard to see how adding "leave" to the mix allows one to discriminate such small groups better than a few "Is this one the Knight/Knave/Joker/etc Troll?".
WHAT ABOUT THE LANGUAGES? SOME SHARE A LANGUAGE REMEMBER?
I remember, and in the real situation this would very probably allow to solve the problem quicker, I just don't see how to do it in a deterministic way in advance. We don't know which three trolls share a language. We cannot deduce from two trolls saying the same thing that they speak the same language; "No" in Madeupish might be "Yes" in Nonsensean. We cannot even be sure we will hear all the possible words in the first question; for example, if 2, 8 and 10 all speak Trollish and all answer "Yes" then we won't have heard Trollish "No". If we have, say a T group with 4 members then we know at least two will speak the same language, meaning there will be two identical "Word1 Word2 Word1", but is it "Yes No Yes" or the Random Troll and Telepathic Troll both saying "No Yes No"? The beauty of the pattern is that it's irrelevant whether a given troll is saying "Yes" or "No" - for example, for the question "Is the Meterological Troll in your group?", the Telepathic Troll might be K-type but giving the opposite answers from the Knight Troll. If it chooses to break the pattern, then two K-types in the other group necessarily had the same pattern and we can go by their answer. If it chooses to maintain the same pattern, thus answering the opposite of what the Knight Troll said, it doesn't matter that its answer was "No" when the Knight Troll's answer was Yes, or that we don't know which is the Knight Troll and which is the Telepathic Troll; we know the real answer is "10 is in the T group" because we know what Knight Troll's answer would be given the pattern, and they both follow the same pattern.
The closest thing to useful I can think of is once we've identified 10, 2 and 3, maybe also 6, and we know exactly what "yes" and "no" mean to each, that might be enough information to tell which of the Invariant trolls has been saying "Yes" and which has been saying "No", thus saving a question to identify them. But I'm not quite sure how to swing it, considering some languages might be the exact inverse of others.
tl;dr: It is entirely plausible, even probable that there is a way of improving this answer based on using the fact that every three troll shares a language, but finding it seems like a lot of work to me. As for whether using that information allows a superior, completely different answer, I can't speak to that.
HOLD THE MOTHERFLIPPIN 'SEND' BUTTON
I have just realized that with the two "are you truthy" questions, and if 4 and 5 try and pass for Truthy which according to the previous reasoning they should (this might change depending on where this goes), I end up with a whopping 7 "No"s on one question and 7 "Yes"es on the other! Victor is right, surely we can do something with this.
1) I am guaranteed to hear all "Yes"es and "No"s after the second question.
2) The distribution of "Yes"es is the following (Trollish=T, Whateverese=W and Nonsensean=N - technically I should call them "language 1", "language 2" or something because I don't know which language is which, but I ultimately don't care. And neither do the trolls, because they're trolls) :
T, W, N or
T, W, N
So we'll hear at least one group of three trolls saying the same thing. Let's assume that "No" in at least one language sounds like "Yes" in at least one other, because these languages are designed to confuse; you know if it isn't true the trolls will even go back in time to change the languages to make it true. Meaning for each configuration, one element will be added to a given language because there were 7 "Yes"es in varying languages and 1 "No" that sounds like a "Yes" in one of those languages. So the configuration will actually look like:
T, W, N.
In which one of the triplets is 3 "Yes"es, and one of the triplets or the double contains a "No" in a different language.
* * * and that's all I've got so far * * *