Paralleled concertina for 2x drive current?Guitar/bass amps generally do not
intentionally draw current from the phase inverter. So you don't need to design for low output impedance and/or current capability for only this reason.
The real driving factor is the grid reference resistor value (grid-to-ground or grid-to-bias supply) permitted for the tube, and how the output tubes are biased.
Check the output tube's data sheet. If you're biasing with cathode bias, the resistance allowed from grid to ground (per tube) is higher than that allowed for fixed bias. That's because excessive resistance could cause a reduction of bias due to residual gas within the tube; gas ions strike the grid, create a slight positive voltage across a too-large grid reference resistor, which increases tube current and liberates more trapped gas ions. Cathode bias is a series resistance which inherently limits tube current in the face of a loss of bias, which is why the allowable maximum grid resistance is higher in this case.
The voltage gain of any tube stage is somewhat reduced by the effect of the load presented by the following stage. You'd like the resistance to ground (or bias supply) of the output tubes to be twice or more the load resistance of the phase inverter. This load resistance is generally around twice the internal resistance of the tube (or higher) to maximize the actual gain of the circuit relative the the possible gain of the tube.
So, if you're using a pair of 6L6's with cathode-bias, the sheet says you should use 500k maximum per grid. Most guitar amps use 220k (or maybe 470k) per tube. With 220k, you'd like to see no more than 100k plate load at the phase inverter, and no more than 50k internal plate resistance of the tube used as the inverter. We get close enough to this with a typical 12AX7 in a long-tailed pair.
If you were using a quad of 6550's in fixed-bias, the sheet says you should use no more than 50k per tube. If we stick to manufacturer's recommendations (and amp makers sometimes cheat this value upward), the parallel pair of 6550's on one side of the OT winding look like 50k ll 50k = 25k. You'd like to have a plate load of 12k (to keep the voltage output from being loaded down by 25k), and an internal resistance of only 6k. That's why the
Ampeg SVT uses the 12AU7-half of the 12DW7 as a split-load inverter with 15k plate load, which is then amplified by a differential pair and coupled to the actual output tubes by cathode followers. The cathode followers are needed to drive 47k ll 47k ll 47k = ~15k load presented by the output tube grids, and the differential driver is needed to provide a little extra boost to make up the voltage loss at the cathode follower and reduce the output requirement on the actual split-load phase inverter.
For your amp, I don't know if you're using 7027's or 6L6GC's (the 6L6's are close enough to the same tube, for a lot less money). The original amp's 7199 triode section is same-as a 12AU7. And Fender's Twin Reverb cheated by using 220k grid-to-bias supply for each pair of 6L6's. I'd say you can use 100k or 220k for each side of the push-pull output section, and use a 12AU7 section in place of the 7199 with no further changes to the stock B25 circuit.