I believe, If I assume ZFC, or rather any other consistent set theory, the answer is :
Such a thing does not exist.
Reasons for the claim:
Let such a thing exist. Call it $C$.
You might call me a number,
but that would be a blunder
It is obviously larger than any natural number as natural numbers are countable and thus bounded by the ordinal $\omega$, and $C>\omega$ because it is "larger than largest" (and so $C$ is greater than any known cardinality, which will form the basis of our contradiction).
I'm in the deck of the cards,
as big as them come.
An allusion to the fact that $C$ is the cardinality of some set, and is very very large (please do not hate me)
You'll never count up to me,
I'm beyond such a sum.
You can only reach me with creativity and wit,
leaving the finite-minded in a rage and a fit!
Since $C$ is larger than $\aleph_1$, we cannot count up to it, rather create hypothetical sets that have such a cardinality.
My innards in size relate,
though you may debate
There is a relation between cardinalities smaller than $C$ and $C$. For example, $\aleph_1=2^{\aleph_0},\ \aleph_2=2^{\aleph_1}$ and so on and $\aleph_0<\aleph_1<\aleph_2$. It is debatable because in ZFC, one cannot prove that there is nothing between these cardinalities.
what is larger than largest,
your intuitions are tarnished.
The final piece of the puzzle. $C$ is larger than any cardinality, which is analogous to the statement that $\omega$ is larger than any natural number. But can $C$ exist?
Claim : No.
Proof : If $C$ exists, then there is a set whose cardinality is $C$. Take the power set of this set. The power set has strictly larger cardinality than $C$ which contradicts the fact that $C$ is "larger than the largest". QED
This actually throws our intuition off track, and we are left dazed.
Hint 2 tells us:
Keep going beyond what we can find. That's what we had done to create $C$ but unfortunately such a construction is not possible as we had seen.
Hint 3 is mathematical :
Burali-Forti is a very well known paradox that states you can't create the set of all ordinals, which seems unintuitive.
Eh?
A mathematical proof is, although not complicated, but involved, and here's how we can understand it. Let the ordinal associated with $C$ be $\Omega$. Now since ordinal numbers are well-ordered in a natural way, this ordering must have type $\Omega$. We know that the order type of all ordinal numbers less than some fixed ordinal number is the ordinal number itself. The, the order type of ordinal numbers $<\Omega$ is $\Omega$. But this means that $\Omega$, being the order type of a proper initial segment of the ordinal numbers, is strictly less than the order type of all the ordinal numbers, but the latter is itself $\Omega$, and we have reached a contradiction.
Okay...
And we know that such a set cannot be formed (at least in ZFC, but in some theories, it forms a class, but that's a different story), and hence has no cardinality. But this should be "larger than the largest". Hence proved.