Energy subtracted from a capacitor for a given voltage drop
You might be better off to use the originator of the energy equation: -

$$Q = CV \rightarrow \dfrac{dQ}{dt} = C\dfrac{dV}{dt} = \text{current}$$

So, if you know the LED current and you know what the initial capacitor voltage is you can estimate dV/dt as a slope taking you from \$V_{INITIAL}\$ to \$V_{FINAL}\$ in so-many seconds (dt). You could make this a bit more sexy by factoring in the dwindling LED current as V falls.