I forgot to tell you to measure d.c. volts on both sides of the 680Ω phase inverter resistor, but no matter. We'll move forward without it.
The output tubes are biased to ~-32v and therefore need a 32v peak input signal (to a single tube) to drive them to full clean output power.
We can probably safely estimate a gain of 30 from the input of the phase inverter to either output. A gain of ~60 would be typical for a 12AX7 stage, and the long-tail is generating 2 outputs with a single input voltage, so gain from the input to either output alone is about half what it normally might be. 32v / 30 = ~1v peak input to the phase inverter to generate full clean output power.
We'll assume the 1MΩ master volume between V3A and the phase inverter is at maximum.
V3A has a 100kΩ plate load, but if the master is full-up, it also sees two parallel 1MΩ pots as a load. Therefore the dynamic load is 100kΩ ll 500kΩ = ~83kΩ. Cheating a little to reduce estimation, the gain of the stage is found by the intersection of the green 83kΩ loadline with the gridlines. Gain = (275v-110v)/(3v-0v) = 55. Therefore, 1v peak/55 = 18.2mV peak input to this stage gives full clean output power.
There is a voltage divider leading into V3A made up of 220kΩ, 3.3MΩ & 470kΩ, with V3A's grid at the 470kΩ. That gives a "gain" (really attenuation) of 470kΩ/(220kΩ+3.3MΩ+470kΩ) = 0.118. So we need 18.2mV peak / 0.118 = 154mV peak at the input of the voltage divider for full output power.
Your Gain switch is off, so V2A is bypassed.
I assumed V1B has a supply voltage (Node A) of ~300v, as that was implied by the tube current, plate voltage and plate load resistor. I also assumed you have a 12AT7 in the socket. The voltage divider at V3A's grid is large enough that there's little error in assuming the dynamic load for V1B is 100kΩ. Gain = (200v-34v)/(4v-0v) = 41.5. Therefore, 154mv peak/41.5 = 3.71mV peak input to this stage gives full clean output power.
Ironically, because the 12AT7 is also used counter to its nature in low-current, high impedance circuitry, it distorts more than the 12AX7 in V3A. In other words, it gives less amplification, but its characteristic curves are so bent down where it's running that it has about double the distortion of a 12AX7 in that spot (assuming small signals).
The Preamp Volume control is not indicated anywhere on the schematic. I'm will assume that it follows the tone stack, like a blackface Fender amp (which stage 1&2 seem to copy anyway). You have your Preamp volume set around 2-3. A true 10% taper log pot set to 3 would only have ~4.8% of its total resistance to ground, which gives a "gain" (loss) of 0.048. 3.71mV peak / 0.048 = 77.3mV peak.
V1A has an internal plate resistance at the operating point of ~23kΩ. 23kΩ ll 100kΩ = 18.7kΩ source impedance for the tone stack (used to tweak the Duncan Tone Stack Calculator model). TSC shows a loss of 11-12dB for lows & highs at the settings in your pictures, and a midrange loss of 22dB.
-12dB converted to a voltage ratio equals -> -12dB / 20 = -0.6 -> 10-0.6 = 0.251, or about 1/4th
-22dB -> a voltage ratio of 0.079, or ~1/12.6
Accounting for the loss of the tone circuit, 77.3mV peak / 0.251 = 308mV peak at Bass/Treble or 77.3mV peak / 0.079 = 978mV at 500Hz at the input of the tone circuit will make full clean output power.
V1A is essentially the same as V1B, so we'll assume the same gain of 41.5. So an output of your pickup of 308mV peak / 41.5 = 7.42mV for Treble/Bass gives full clean output power, or 978mV / 41.5 = 23.6mV for 500Hz.
The average pickup, whacked hard should make 100mV pretty easily, and turning your guitar volume to half still gives 10mV, so you see even these settings cause the output tubes to distort. All preamp tubes have distorted some as well (I found 2nd harmonic distortion between 2-10% for various stages) even before they're hit hard enough to either cutoff or saturate.
From what I've found, your distortion issue is "normal" for the circuit in the STP-G. In fact, I'm almost certain distortion was the whole point.