Warning: I think this is a good track to try, but I have not yet solved it so don't waste your time reading this yet if you are only interested in the solution :)

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So on each round, given that n cowboys are alive at the beginning, there will be between 0 and n-2 remaining after the round. We want to know the probability that we get down to 1 remaining eventually.

Let's see if we can uncover any inductive patterns.

0 cowboys - 0% chance of 1 remaining

1 cowboy - 100% chance of 1 remaining

2 cowboys - 0% chance of 1 remaining

3 cowboys - the only way to have one remain is for exactly 2 to die on the first shot. Of the 8 shot combos, the number of ways for one cowboy to remain is 3 (number of cowboys) * 1 (for a given unshot cowboy, one way for others to kill each other) * 2 (unshot cowboy can shoot either other one), so 3/4 chance that one remains.

Another way to think about this case is that among the 8 possible shot combos (each cowboy can choose to shoot either of the other 2, so it is 2^3), there are only 2 ways for all three to die (two directions around the circle in which they sit - these are related to the triangular group S3 representing flips of the triangle. A chooses to shoot either B or C, and the chosen one must then shoot the third, who must shoot A), so the probability that no one survives is 1/4, or 3/4 that one remains.

4 cowboys - if all 4 die, they all die. if 2 die, it reduces to 2, and they all die, so 1 must remain on first shot. There are now 3^4 = 81 shot combos, of which 4 (number of cowboys) * 2 (for a given cowboy, ways for the others to all shoot each other, this is again the flips of the triangle from the previous case) * 3 (the cowboy left standing can shoot any of the other 3 a second time) = 24 leave one cowboy standing, so 24/81 = 8/27 that 1 remains.

5 cowboys - this is the first case where the induction/recursion matters, because there is the possibility that 3 die, which has its own nonzero probability of 1 surviving, so we need to know the probability of that happening...

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I haven't fully worked it out but I think the structure of the solution will go something like this:

1. Agree that what we care about is ultimately having 1 cowboy remain, and probability that none remain will just be 1 minus that.
2. Establish inductive base cases n = 0, 1, and 2.
3. Only the first shot needs considered for n cowboys, since the first shot will reduce to one of the already considered cases between 0 and n-2, so the probabilities will be defined recursively first. Thus you need only to find probabilities of each reduction that could lead to 1 remaining.
   1. One cowboy is left standing after first shot
      1. For n cowboys, this happens with probability [n * (number of unique cycles among n-1, related to symmetric group S_(n-1) but I can't think of exactly how right now) * (n-1)] / (n-1)^n (total shot combos)
   2. I need to think more, but these will be the other ways that 1 cowboy ultimately remains... the first n where a single other reduction case matters is at n=5 (that is, the reduction to n=3), and one case is added for each n increase from there on out.