Suppose the cylinder is very wide. Then certainly centrifugal force would cause the fluid pressure to be higher at the perimeter than it is at the center. So if the hole is near the periphery, there is a greater "head" there, so fluid should be ejected at higher velocity.
Ignoring viscosity, the velocity should be proportional to square root of pressure.
EDIT: Got some drive-by downvotes, but here's the mental picture of the rotating tank.
Anyway, in the $\omega = 0$ case, the pressure at the orifice is just proportional to height of the liquid above it.
In the $\omega = high$ case, assuming the liquid is rotating at the same speed as the tank, then the radial acceleration of a unit of liquid is $r\omega^2$, which has to be integrated over the radius of the liquid from rim to surface, and multiplied by density $\rho$ to get pressure.
Clearly that can be very high.
Then, if you want to know the ejection velocity, Bernoulli (much simplified) says $V^2 = 2p/\rho$.
I'm not sure but I suspect in the limit the ejection velocity would equal the rim velocity (just like throwing a baseball).