First, observe that:
the number of sides of gloves is exactly equal to the number of people that must touch at least one side of at least one glove.
by the "pigeonhole principle", with exactly $n$ gloves, each side of each glove must be reserved exclusively for being touched by at most one person before it is irreversibly contaminated.
During the procedures a surface of a glove can be in one of 3 states:
"clean" - i.e. Not touched by anyone, and only brought into contact with other "clean" gloves.
"used" - i.e. it has been touched at least once by the person two whom it is allocated, and has never been brought into contact with a glove that is not clean.
"contaminated" - i.e. it has been touched by someone other than the person to whom it is allocated (not allowed!), or brought into contact with another used glove (allowed, but this glove can now never be touched by anyone).
If it's possible to generalise to $n$, it must necessarily be possible to do 3, so I'll start with that simpler case, where doctors X, Y, Z must all operate on patients A, B, C.
Without loss of generality, X performs the first operation on A. There are two possibilities:
A single glove gets touched by X on one side and A on the other. This single glove must then be worn by X for the operations on B and C (with at least one other glove to protect B and C from A), and also must be worn for the operations on A by Y and Z (with other gloves inside to protect Y and Z from X). This glove will contaminate any other glove it touches, so Y and Z must complete both operations on B and C before this glove is used again, leaving non-contaminated used surfaces on both gloves. Unfortunately the known solution for $n = 2$ results in two gloves becoming contaminated on one side, so this does not work, and furthermore, even if that were possible, whichever of "Y operates on A" and "X operates on B" were to happen first would contaminate the other side of the "X/A" glove.
X wears two gloves to operate on A, one inside the other. This results in one glove "used by X" on one side, and another "used by A" on one side. In a different variant of this puzzle, where each operation uses two gloves, other doctors can then use these gloves to operate on other patients by turning one of the gloves "used by X" inside out, so the two "used by X" surfaces face each other, without contaminating any other glove. However, in this puzzle, the next person who uses the "doctor X" glove must have another glove outside, which becomes contaminated on that side, so must have been used for all 3 of its operations first, and will thus contaminate the X side of the glove. Therefore X must complete all 3 operations before the other side of that glove is touched. However, a similar argument shows that all 3 operations on A must be completed before the other side of the A glove is touched. That only leaves 1 glove, which is not sufficient to allow X to complete 2 more operations AND for 2 more operations to be performed on A without at least one side of at least one glove becoming contaminated - the remaining glove must be touched by at least one doctor (performing the second operation on A) and at least one patient (for the second operation by X) both whilst the other side is "clean". Contradiction.
It is thus clear that
$n$ gloves are not sufficient, and at least $n+1$ gloves are necessary.
My initial intuition was that
$n+1$ gloves are sufficient, with the additional glove having one side kept "clean" at all times.
So I attempted to come up with
a procedure to allow 3 doctors X Y Z to operate on 3 patients A B C using 4 gloves, in a manner that would generalise to $n$ doctors using $n+1$ gloves.
The first part was relatively easy - e.g.
X and Y uses gloves 1 and 2 respectively to operate on all patients except A (with an allocated glove for each patient) keeping the outside of their glove clean and the inside of the other patient's glove clean, then Z operates on C and B using those gloves, and X and Y operate on A using the glove that's still clean on the outside, and finally Z operates on A, re-using any pair of gloves that were "used by Z" and "used on A" (with the "used on B/C" side of one glove touching the "used by X/Y" side of the other).
For the second part,
I tried various ideas, such as having X's glove turned inside out and placed inside Y's glove after Y had operated on A, so that a "clean surface to A" interface is available, but either got confused, or ended up with procedures that wouldn't extend past 3.
Thus it seems that
my known procedure so far generalises to $2n-2$ rather than $n+1$... i.e. 2 gloves for n=2, 4 for n=3, 6 for n=4, etc.
This remains a "partial answer" while I
further check for ideas that might apply for the case $n=4$ to see if this upper bound can be improved upon, or find some insight towards a proof that $2n-2$ is always needed.
The $2n-2$ solution only results in two sides of gloves becoming "contaminated", and my intuition is that an optimal solution could contaminate anything up to both sides of all gloves except the ones used in the final operation - giving plenty of scope for further improvements.
A later refinement of that intuition while I was considering improvements to @vepir's answer which I never got completely nailed down was that
there is never any need to contaminate the second side of any glove, as the operation that contaminates the second side would need at least two other gloves, and could equally occur without that glove used at all (allowing the contaminated surfaces of the inner and outer gloves to touch directly), so an optimal solution would end with all gloves contaminated on exactly one side. This is precisely what @loopywalt's answer does, using a method I'd started to investigate, but not followed to its conclusion, so I've got good reason to believe that one optimal.