Electronic – Supercapacitor with a rated voltage value greater than 2.7V feasible

The problem is the fixed 15v output voltage of your DC-DC converter. Capacitors should be charged with a current output, not a voltage output.

If the voltage on a capacitor is less than the output voltage of a power supply feeding it, it will take as much current as the supply cares to supply, in other words it behaves like a dead short.

The DC-DC converter can still be buck, can still use the same power stage, but the controller should be changed to control the maximum current output to safe levels, as well as the maximum voltage to be safe for the supercaps.

With regenerative braking, ideally you would program the converter to put a braking torque on the wheels proportional to the position of the brake lever/pedal. This is naturally what you would get if you try to mimic the response of how mechanical brakes would behave, I presume you want the electric system to behave as a driver would expect it to behave.

It seems that your either your calculations are far off, or you have a bad battery pack - particularly, you have at least one weak cell.

This document suggests that, to get 11.5 volts during peak discharge you must be drawing on the order of 2C. For a 300 Ah battery that's 600 amps+. Of course, different batteries have different discharge characteristics, so the curves should be taken with a grain of salt, but I'm inclined to accept them as a starting point.

As for using a supecap, the calculations are pretty straightforward. $$\frac{dv}{dt} = \frac{i}{C}$$ or in simple terms a 1 F supercap will discharge at 1 volt/sec for a 1 amp current. Even assuming your current peak is only 100 A, you'll get a 100 V/s droop per farad. To produce a 0.5 volt drop in 400 msec, you'll need $$C = \frac{i}{\frac{dv}{dt}} = \frac{i\times dt}{dV} = \frac{100\times 0.4}{.5} = 80F$$ or about 27 of your 3F units, for a cost of about 1350 dollars.