Do the voltage changes depend on the load? If so, what is the load of the LED?
The "capacitive dropper" circuit is approximately a constant current source, not a constant voltage. If you change the load, then the voltage across the output will change.
I looked up in the datasheet that the LED requires 3.3 V forward voltage and 300 mA to make it to 1 W of power. What will happen if I use 3.3 V with 130 mA? Will the LED turn on?
LEDs will work at well below their rated current. They will still light even at 1mA, perhaps even 0.1mA. But you can't "use 3.3V". The current through the LED determines the voltage it will drop. It might be as low as 2.8V at very low currents.
What will happen if I add a 3.3 V Zener diode parallel to the LED? Will the current change?
You have 10 LEDs there. Putting a zener across one of them might drop the current through the LED a bit. But I can't think of any reason why you would want to do it. If you put the zener across the whole string of 10 LEDs, they will all go out.
But whatever it is you're trying to do, a zener diode probably isn't the answer.
This looks suspiciously like it's an "X-Y question". An X-Y question goes something like this...
- I want to do X
- I don't know how to do X
- If I could do Y, then it might help me to do X
- I don't know how to do Y, either
- I will ask someone how to do Y.
See The XY Problem
Big hint: If you want to change the current through the LEDs, change the value of capacitor C1. It's that capacitor that determines the current that flows through the whole circuit.