Let the resistance of a single piece of wire connecting the drivers be R. Let the current to each driver be the same, call it I. Assume it's steady state, which it's nowhere near so pretty poor assumption...Also assume no impedances other than the wire resistance ... again we should really include it but there's not really enough information to make it worthwhile so I didn't bother.

1st driver gets the supply voltage, V.
2nd driver: V-2*3IR = V-6I
3rd driver: V-2*3IR-2*2IR = V-10IR
4th driver: V-2*3IR-2*2IR-2*IR = V-12IR

So as you can see the last driver is in trouble if I and R are significant. The good thing with your drivers is they're at mains voltage, so I is not very high (maybe an amp, could peak at more) and R is (assuming 1.25mm^2 wire) about 14mOhm. That means you're loosing about 0.2V to the last driver. Compared to the 240V that's not much, but maybe when the currents are changing quickly it could be a problem as then the impedances I ignored would have an effect ... and all the drivers will be drawing different currents.

If you have 50V drivers then a reasonable average current might be 4 amps. Lets say you used the same wire, that's 50-12*4*0.014=49.3V
The thing is that DC voltage will now fluctuate, there's now about 1% ripple on the power supply. This could cause a problem if the smoothing capacitors are not big enough in the driver.

As m_c said, there's more to it than just the voltages.

(On reflection I guess working out the above didn't achieve much, but I thought it was interesting!)