You seem to have understood my maths perfectly, but you missed where I said "about" for the capacitor value.
The quick calculation is to simply ignore the LED voltage drop. The impedance of the 2 µF capacitor is 1326 ohms as you worked out. That will limit the current to 90 mA.
Where the voltage drop across the capacitor is close to the supply voltage, then you can't calculate the voltage across the LEDs with any accuracy.
As you wrote
sqrt(120^2-119^2)=15.5
but if the current is just marginally less, the voltage across the capacitor falls very slightly. Let's just assume it drops to 115 V
sqrt(120^2-115^2)=34.3
As you can see, the LED voltage has doubled but the voltage on the capacitor has only fallen slightly.
If the voltage drop on the LEDs, resistor and rectifier is 65 V, then the voltage left on the capacitor is
sqrt(120^2-65^2)=100.9 V
That would indicate that maybe an increase of the capacitor to 2.4 µF would be needed.
I don't think this analysis is perfect. Taking the sum of the squares is only fully valid for capacitors and resistors, and the voltage drop across the rectifier, resistor and LEDs is not sinusoidal.
What it all comes down to is that 2 µF will limit the current to around 100 mA DC, and that won't be reduced much by the voltage drop on the LEDs.
With the large voltage drop across the LEDs, I am not entirely sure about the details.