Killing Kickback with capacitors
And how do i calculate the amount of current/energy, that will go into
  the capacitor?


A current of 50 amps through an inductance of 3.2 mH implies a total stored energy of 4 joules (\$E = LI^2/2\$). If this energy were wholly dumped into one of your 1000 uF capacitors across the rail, the voltage would rise according to the energy formula for a capacitor (\$E = CV^2/2\$).

So equating 4 joules to the above gives a voltage rise of 89 volts. But this is when the initial charge on the capacitor was zero.

If your capacitor is already charged to 600 volts, it already has an energy of 180 joules and 4 more joules makes 184 joules or, a final voltage of 607 volts after the inductor energy has dumped.

However, if your capacitor has an ESR (equivalent series resistance) of 0.1 ohms, the 50 A that would initially flow into the capacitor would also produce 5 volts across the ESR so the peak of 607 volts could be as high as 612 volts.

Is this value OK? Only you can say. I would also recommend you download a free spice tool (such as LTSpice) as this can help a lot and is worth the up-front learning curve for anyone in electronics.