How does the values of the capacitor affect the filter circuit?
The smaller capacitors will not pass as much current at 60Hz, so the bulb will not light (or glow as brightly)  

We can calculate the expected wattage:

For 1uF:
1 / (2 * pi * 60 * 1e-6) = 2652 Ohms.  

If we add this to the bulb resistance (say 50 ohms as it will be somewhere in between cold and hot) we get 2700 ohms (we can be rough here as the bulb resistance makes little difference)    

For a 115V line, that's 115 / 2700 = ~43mA.
So the apparent wattage will be around 115V * 43mA = ~5W
The real wattage dissipated by the bulb will be much less, around 100mW.
We calculate this by I^2 * R. So with a 50 ohm filament resistance we get 0.043^2 * 50 = 92mW.    

Not enough to light the bulb up. For the 2.2uF the result would be around 900mW (may glow a bit?)   

If we do the same for the 22uF capacitor, we get:
1 / (2 * pi * 60 * 22e-6) = 120 Ohms.
sqrt(120^2 + 285^2) = 309 Ohms.
115V / 309 = ~372mA
115V * 372mA = ~43W  (~39W dissipated by filament)

So you see the larger cap makes a big difference.