Great question :-)
Let $N$ be the event that no-one was killed, and $C$ the event that Aidan is crazy. Then you want the conditional probability
\begin{align}
\textsf{Pr}(C\mid N)
&=\frac{\textsf{Pr}(C\cap N)}{\textsf{Pr}(N)}
\\
&=\frac{\textsf{Pr}(C\cap N)}{\textsf{Pr}(C\cap N)+\textsf{Pr}(\overline{C}\cap N)}\\
&=\frac{\frac12\cdot1}{\frac12\cdot1+\frac12\cdot\frac13}\\
&=\frac34\;.
\end{align}
So the chance wasn't quite as high as you thought, but it was still the right choice.
You calculated
\begin{align}
1-\textsf{Pr}(\overline{C}\cap N)&=\textsf{Pr}((\overline C\cap\overline N)\cup(C\cap N)\cup(C\cap\overline N))\\
&=\textsf{Pr}((\overline C\cap\overline N)\cup C)\;,
\end{align}
the unconditional probability that either Aidan is Mafia and someone was killed or he's crazy (and hence no-one was killed). I don't think this corresponds to any conditional probability given your knowledge of $N$.
Regarding your question "how can it be a $50\%$ chance he's a crazy person, but actually an $80\%$ chance" (in fact $75\%$), the former is the unconditional probability for him to be a crazy person and the latter is the conditional probability, given your knowledge of $N$.
(By "unconditional", I mean conditional only on the facts that you regarded as definitely settled: That the three of you were all townspeople and Aidan was either Mafia or crazy. Of course the original unconditional probability for Aidan to be crazy was only $\frac15$.)
Assuming no one ever respins the chambers (because otherwise the strategy is trivial—always pass the gun off):
Work backwards. Obviously, with only one chamber remaining, you would pass the gun off. Good luck captain!
With two chambers left, if you pass the gun, you have a $1/2$ chance of surviving, no matter what you do (pass the gun, or shoot again).
With three chambers left, if you pass the gun, your opponent has a $1/3$ chance of surviving that shot, after which your chances are both $1/2$. So your chances of survival by passing the gun off is $2/3$. So you pass the gun off.
With four chambers left, if you pass the gun, your opponent has a $1/4$ chance of surviving that shot, after which they will pass the gun off, and your chances of survival at that point are $1/3$. So the chances are $1/2$ for both, and it doesn't matter what you do.
With five chambers left, if you pass the gun, your opponent has a $1/5$ chance of surviving that shot, after which their chances of survival are $1/2$. So your chances of survival if you pass the gun off are $3/5$, so you pass the gun off.
You should be able to show by induction that passing the gun off is never a bad choice. With an even number of chambers left, it's always a $50$-$50$ proposition, and with an odd number of chambers left, it's strictly better to pass the gun off.
There might be an analysis that short-circuits the induction and makes the pattern obvious upfront, but I don't see it yet.
Best Answer
For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing. Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from)
Player 1 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$
Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$
Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$
Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$
Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$
Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$
So the two player game is fair without shuffling. Similarly, the $3$ and $6$ player versions are fair.
It's the $4$ and $5$ player versions where you want to go last, in hopes that the bullets will run out before your second turn.
For a for $4$ player game, it's:
P1 - $\frac26$,
P2 - $\frac26$,
P3 - $\frac16$,
P4 - $\frac16$
Now, the idea in a $2$ player game is that it is best to be player $2$, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player $1$ (or your captor), thus winning, changing your total odds of losing to P1 - $\frac36$, P2 - $\frac26$, Captor - $\frac16$