In simplified terms, the governing equation for a “perfect” capacitor is:
I = C dV/dt
which states that the current is equal to the capacitance times the change in voltage with respect to time. In layman's terms, you can get a first approximation like this:
I = C (Vinitial – Vfinal)/time
Example, If:
C = .47F
Vfinal = 5.0 Volts
Vfinal = 3.3 Volts
time = 7200 seconds
the capacitor would be able to supply
I = .47 (5.0-3.3)/7200 = .000111 amps = .11mA during that time frame (the voltage would change from 5.0 V to 3.3 V after 7200 seconds while supplying the shown current)

However, the equation works both ways – if your circuit draws more current, for example .001 amps, you can determine how long you can keep your capacitor voltage at 3.3 or above like this
I = 0.001 = (5.0-3.3)/time
solving for time gives time = 1700 seconds (about 28 minutes).

I picked a lower limit of 3.3 V – you need to determine this for your setup. Also, with the nanoWatt controllers, you should be able to get your current consumption much lower, like several uA or less, which means you could have a backup for days.

Last, no system is perfect – there are leaks and losses and other things going on that you need to be aware of – design your power backup to last longer than you really need it – and test it!