Why do MOSFET drivers need so much current?
A current of the gate charge *switching frequency is consumed from the supply. With 6 FETs, this is then 6*350nC*25kHz = 53 mA. This generates a power of 53mA*40V=2.1 W. This power is dissipated between the IC and the gate driver resistors. That total power can be calculated without knowing the exact waveform shape. 

To know how much is in the gate resistors vs. the IC, you would have to know the driver on-resistance of the IC, and then the power is shared in proportion to the resistances. The A4935 has gate driver resistance of about 10 ohm, and you have about 5 ohm gate resistors, so 67 % of the power is in the IC, and 33 % is in the 6 gate resistors, so the IC dissipates 1.4 W, and each resistor dissipates about 0.1 W.

'Too hot to touch' is about 70-80 C for a plastic package. Depending on your PCB heat sinking it is possible that a temperature rise from ambient (25 C) is caused by 1.4 W.