It would take about 2 ms to charge 10 µF from 0V practically all the way to the battery voltage. The capacitor wouldn't ever fully discharge though, so 1 ms would be a closer estimate.
If the PIC goes into sleep mode, its average current could be lower than 10 µA, using less than 50 µJ per second. The MOSFET gate, once charged, requires no current; its capacitance is around 5 nF so if charged from a PIC output with no gate resistor, it will charge in under 10 µs requiring less than 0.5 µJ so that's not significant compared to the PIC's load.
If we recharge the smoothing capacitor 10 times per second, at 10 µA load current and allowing for a 1V drop in VDD voltage every 100 ms, the reservoir capacitance only needs to be 1 µF.
If the resistor is increased from 33Ω to 330Ω a 1 µF reservoir will take about 1 ms to recharge and must be recharged every 100 ms. So the LED duty cycle would drop from 100% to 99%.
So here's my suggestion. Use 1 µF for the reservoir capacitor on your switch board, and use 330Ω across the LED driver to provide the top-up current for that reservoir capacitor. Have the PIC switch the MOSFET OFF for 1 ms every 100 ms. I believe that should work safely. My calculations have all been conservative.
I specified a separate decoupling capacitor (CD) and reservoir capacitor because I was thinking the reservoir capacitor would be a tantalum or aluminium electrolytic, which is not suitable for decoupling. But if the reservoir capacitor is only 1 µF you could easily use an X7R ceramic and CD wouldn't be needed.