Expanding the capacitive divider

Remember the sine wave is generated by the rotating vector. The amplitude of the peak can be determined as the sine of the angle at any given instant.

The voltage across the components will not add to the value of the input at any given instant because they are not in phase with each other. When we add the signals the using the vector they will equal the input. When we plot the signals the result is a right angle triangle. We can use HYP = SQRT( A^2 + B^2) to solve for the values.

Now back to this one. The load and Xc are equal and producing equal voltage drops. Because of the phase angle they will not add to the supply.
Ec = 170
Ert = 170
Et = SQRT(170^2 + 170^2)
Et = 240
I = E / R
I = 170 / 800
I = 212.5 ma
Back to our original circuit. We need to know the number of tubes, their voltage rating, and their current rating. As long as we can make a string with the same current we can produce a working circuit. We could use a 300ma tube in series with 2 150ma tubes in parallel. We would transpose the formula to calculate the cap value.

EXc = SQRT( Et^2 - Ert^2)
Xc = Exc / It
etc
The original circuit would work as long as the load voltage was low. As the load increases the phase angle increase and Zt increases reducing the current.  With one or two tubes it would be close enough. In the last example it is 212ma. The tubes may still function but at reduced capacity.

Tip of the hat kiddo. I have an old radio in need of a panel light. The 15 volt lamp in my junk box may find a use.