Using capacitors to prevent surge current on lead-acid batteries?
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.