Output Capacitor Voltage Holding time during dropout
Let's try to keep it simple by making some assumptions:

The formula, as quoted, is C = I*dt/dV. 
Assuming the average input current is 0.25A, the voltage drop is 4V (i.e. from 12V to 8V) and the time is 10s, then C = 0.25*10/4 or 0.63F. Note that this is the MINIMUM value which will meet this requirement, and that it ignores the probable extra current demand from the converter as the input voltage drops. The efficiency of the converter is also a variable. Furthermore, you need to isolate the input from the capacitor during the dropout time - a diode will do the job, with a slight loss in overall efficiency.

In short, the suggested value of 1F appears reasonable. Have you considered a backup battery instead?