Replacing battery by supercapacitor
You need four pieces of information

1) How much current do you need (in amps)?

2) How long do you want to run your circuit from the cap (in seconds)?

3) What it the largest voltage (normally the regular operating voltage) which will be applied? and

4) What is the lowest voltage out of the capacitor which will still allow the circuit to function?

Let's call 1) i, let's call 2) $\delta$t, 3) is Vmax, and 3) minus 4) is $\delta$V.

The relationship between time, voltage and current in a capacitor with a value of C in Farads is $${\frac{i}{C} = \frac{dV}{dt}}$$
For a constant i, which is a reasonable first approximation in this case, this becomes $${\frac{i}{C} = \frac{{\Delta}V}{{\Delta}t}}$$, or$${C = \frac{i{\Delta}t}{{\Delta}V}}$$ and the capacitor must have a voltage rating of Vmax.

To walk you through this, lets say you need .25 amps for 5 minutes (600 seconds). Let's say your battery puts out 3.7 volts when fresh, and your circuit will work down to a battery voltage of 3.0 volts. Then $$C = {\frac{(.25)(600)}{(3.7 - 3.0)}}$$ $$C = {\frac{150}{0.7}}$$

and you need a 214 Farad supercap with a voltage rating of 3.7 volts minimum.

EDIT - As Brian Drummond has pointed out, supercaps often have a high internal resistance. If you want to take this into consideration (and you had better do so), you need to quantify the resistance R of the cap at the current level which you are using. Then the capacity calculation remains the same, but if the capacitor voltage rating is Vcap, instead of being Vmax, $$Vcap = Vmax + (iR)$$ In the above example, if the ESR of the supercap is 20 ohms, $$Vcap = 3.7 + (.25)(20)$$ and $$Vcap = 8.7$$ In this case you would definitely need to find a different supercap.