Raspberry pi super capacitor power supply
The relevant formula for this sort of circuit is below:

$$C_{min} = \frac{t_{HOLD}\cdot I_{OUT}}{\Delta V}$$

In your case, neglecting the voltage drop of the current-limiting resistor, and assuming you can run down to 3.3V, this becomes

$$C_{min} = \frac{10\text{ s}\cdot I_{OUT}}{5\text{V} - V_{diode} - 3.3\text{V}}$$

Ballpark numbers:

Use a Schottky diode, and you'll maybe have 300 mV for the diode drop. Let's say the Pi draws 0.4A.

With these numbers, the equation above gives 2.9F. So your capacitor is close, but might not be quite enough.

A 12V battery, charger, and buck converter might be a better way to do this. You'd at least be able to avoid the giant capacitor.