There are several approaches to this problem, but I like to have the tubes as active elements in the flasher.
From the transfer curves provided by @Olin, it seems that there is gain available at 27V, but the total current is limited to less than the typical forward junction current of a LED. It should still be visible.
The simplest circuit would be to put the DC supplies in series to make a 32v B+ supply. The tube heater is connected across the 12V DC supply.
Here is a schematic that may work. It should be a good place to start.
Some notes, the 5v supply is used to assure negative grid voltage when the 555 output is low. The resistor on the output of the 555 (with the missing value) should limit the grid current, and is probably in the 50K range.
Ra, Rb, and C should be set to give you the pulse rate and duty cycle you want. 555 application notes include formulas for calculating these values.
The screen grid and suppressor grid are connected to B+ to encourage a few more electrons to reach the plate.
There is not resistor to limit the LED current. The tube conductance should be a strong enough limiter.
If you want more brightness, use two 6BA6 tubes with the filaments wired in series (to make up the 12V supply) and all other elements wired in parallel.