If you say 100Hz and 10Vpk minimum (Motor Volt = Freq * KV) and 7V min for 7805 operation, at I=1A, capacitance must be C = 1/(2*100*3) or 1,667uF. If you're including diode drops, then the Vpk might be 8.5V and Vmin 7V, giving 3,333uF.
I'd double each of those and round to nearest highest E12, to give some extra margin.