Let's consider Player B and their chance of winning as a function of a, Player A's roll.
There is an 1−a chance B's initial roll exceeds A's, allowing them to outright win. Otherwise, there is an a chance that B needs to roll again, and then there is a 1−a chance that B exceeds a without exceeding 1. This makes B's chance of winning a(1−a)+(1−a)=(1+a)(1−a)=1−a2.
Now consider Player A. Given their first roll being a, they have a choice to roll again, depending on whether it increases the odds of winning. In the first scenario, there is a 1−B(a)=a2 chance of winning. In the second, the new roll is uniformly distributed from [a,1−a), though we can only count rolls below 1. So there is a ∫a1(1−B(x))dx=∫a1x2dx=31x3x=a1=31(1−a3) chance of winning.
So between these options, we want to choose the maximal value for any initial value of a. Since this is a uniform distribution, this makes the expected value ∫01max(a2,31(1−a3))da.
Between the two expressions in the max function, there is a value of a at which the first expression will strictly exceed the second. It works out to a≈0.532089 being the root, making ∫01max(a2,31(1−a3))da≈∫00.532089a2da+∫0.532089131(1−a3))da≈0.453802.
Hence, A's chances of winning are about 45.38%.
Answer: 45.38%
Extra Credit
As expected, things get more complicated with Player C in the fray. Similar to the first part, Player C's chances of winning, as a function of a (Player A's roll, if legal) and b (Player B's roll, if legal) is 1−(max(a,b))2. For the remainder of the problem, we can denote this as C(x), where x is simply the maximum legal roll between Player A and B, or 0 if neither are legal.
Player B has an a chance of needing to spin twice, contributing a a∫a1(1−C(b))2db=a∫a1b2db=31a(1−a3) chance to win. There is a 1−a chance of having the choice to spin again, contributing a (1−a)∫a1max(b2,∫b1x2dx)db, which we know from the first part to be about (1−a)(∫amax(a,0.532089)b2db+∫max(a,0.532089)131(1−b3))db)=31(1−a)(41max(0.532089,a)+max(0.532089,a)3−max(0.532089,a)−a3+43) (final integral calculation).
Finally, Player A has to defeat both Player B and Player C; Player A also has a choice of whether to spin again or not. Without spinning again, the odds are (1−B(a))(1−C(a) for a given roll a, which works out to (1−31(1−a)(41max(0.532089,a)+max(0.532089,a)3−max(0.532089,a)−a3+43))a2. When rolling again, we actually integrate this expression from a to 1 to get ∫amax(0.532089,a)(1−31(1−x)(40.5320894+0.5320893−0.532089−x3+43))dx+∫max(0.532089,a)1(1−31(1−x)(4a4+a3−a−x3+43))dx. Honestly, the integrals were getting too verbose to keep typing, and I was running low on time, so I plugged it all in here. Not sure if it's right, but we got 0.489523. Seems suspicious, since I would expect lower chances with more players.