Well it's 5V through a 56 ohm resistor, so the charge current will of course vary depending on the capacitor voltage. When the keepalive first starts discharging, the 'recirculated' current will be essentially zero, and will increase as the cap discharges.
But for easy figuring, let's assume a worst case constant charge current into the cap at 2.5V (pretty typical for the 2.5F version at the end of a long boost). That means the power dissipated in the resistor (the main source of energy loss) is 110 mW. If we assume, again worst case, a 10 second run time, that's 1.1J energy lost. From the energies tabulated in my other post, the 2.5F version stores 26.3J. So that's ~4%, as a worst case bounding value. The actual loss will be less.
It would be elegant if we had a constant current charge regulator that could charge at a fixed 100mA or so. But, it would have to be small, run off a wide range of input voltages, and be configurable for the required charge current and cutoff voltage. Then the loss would be only the inefficiency in the regulator itself. But unless the implementation of that was as cheap and simple as the regulator, I wouldn't trade the complexity for a couple percent efficiency. Just use the 5F version if you have some monster that needs more runtime, lol.
