I agree with the above poster who said to find out where / why you measured 108V.
Also check connections are good.
Did you measure the 108V when fan was humming and stalled (in a locked rotor condition)? Basically, the large locked rotor current would cause a voltage drop in the wiring from the panel so that the 120V (124.5V) is dropped to 108V. (Knowing circuit length and wire gauge, one could calculate how much current it takes to drop 16 volts (or whatever it is) in a #12 AWG wire).
I’m surprised the breaker doesn’t trip. Maybe it would eventually. It’s probably a race between the breaker and the motor making smoke because it’s wires can’t handle large locked rotor currents for very long before burning up.
Motor starting capacitor may be toast.
For giggles let’s say circuit’s length from panel is 100’, so 200’ total travel in hot and neutral conductors. Say #12 gauge wire has resistance of 1.93 ohms/1000’. So 200’ of wire would be .386 ohms. 16 volts drop through .386 ohms only happens when 41 amps is “humming” through the fan.
As a rule of thumb, the “locked rotor” (aka “stalled”) current of a motor is about 6 times its normal current. So 41 amps is in the ballpark.
If you look at the trip times (curves)of a 20amp circuit breaker experiencing 41 amps, I bet your breaker would of tripped right on schedule, probably around 10 seconds.
If you had wires crossed and a dead short, breaker would of tripped in about .01 seconds.