Calculating Capacitor Discharge Time
Basic capacitor equation:

\$ Q = V C \$ (stored charge = voltage * capacitance)

Current is the time derivative of charge:

\$I = \frac{dQ(t)}{dt}\$ (current = change in charge over change in time)

The time derivative of the capacitor equation is:

\$\frac{dQ(t)}{dt} = C \frac{dV(t)}{dt}\$

\$I = C \frac{dV(t)}{dt}\$

Making a coarse assumption that your discharge current is constant (at 1A), your voltage will decay linearly with a slope related to the capacitance.

That is to say, if your capacitor voltage starts at 5V and you use a 5F capacitor and you draw charge away at 1A:

\$1A = 5F \frac{dV(t)}{dt}\$

\$\frac{0.2A}{F} = \frac{dV(t)}{dt}\$

so your voltage will decay at a rate of 0.2V per second, and you will hit 4.5V in about 2.5seconds...

In order to decay 0.5V over 15 seconds plug in numbers like this:

\$1A = C \frac{0.5V}{15s}\$

\$C = 1A \cdot 15s / 0.5V = 30F\$

You can always put more capacitors in parallel to achieve an additive capacitance.