Okay, then. Counting on fingers...
It seems to me that a good rough guess for the load imposed by the tone circuit is to look at the slope resistor; it's 100k in this case. If the wiper of the mid pot is at the "top" in the schematic, then the 100k resistor is grounded by virtue of the 0.022uF cap. So I seeing this as, best-case, a 100k load on the previous 12AX7 stage.
A "real calculation" involves plotting the loadline of the 12AX7 stage given the cathode resistor and supply voltage, then finding the actual internal plate resistance (and mu, cause it's not exactly constant) at the operating point. But we'll simplify by assuming 62k internal plate resistance and a mu of 100.
Taking those numbers, the actual amplification of the gain stage is dependant on the internal resistance of the tube and the parallel combination of the plate load resistor and external load. 100k ll 100k = 50k, and so that's the total load the tube sees.
Amplification = mu * [50k/(50k+62k)] = ~44
So that's all we can expect from the input stage (but we're not done yet).
What is the bias on the output stage? I skipped over seeing that we're probably talking about a SE output stage, but the comments still apply. For 6L6, EL34, KT77, we won't be as high as the -45v or so needed in a higher-voltage class AB amp, but it would be likely to have a bias of as much as -25 to -30v. The grid reference resistor will likely be ~220k (though in some cases you could use 470k). We'll figure on the lower value.
Again, assuming 62k internal resistance, a mu of 100 for the second gain stage, and 100k ll 220k = ~67k, we get:
A = 100 * [67k / (62k + 67k)] = ~52
To comfortably drive our estimated output stage (we should verify what the true bias and supply voltage will be), we need 30v peak / 52 = 0.58vpk.
We could now divide by the estimated 1st stage gain, but we're forgetting something... you need a volume control. You want to be able to use the volume to compensate for various pickup types and outputs, and you want it to distort before the volume is full-up. By the time you turn up a 10% audio taper pot to "7" it is only at 20% of it's total resistance. That means the signal out is 1/5th the signal in.
So, 0.58vpk * 5 = 2.9vpk
We have not assumed yet the loss presented by the tone stack, we simply figured how hard it would slug the gain of the first stage. As quick guess based on the Duncan Tone Stack Calculator, with pots set midway (and ignoring the need to figure dB based on specific input and output impedances), the tone stack will likely result in about 16dB of loss. That means 1v in gives 0.025v out, which is 1/40th the input signal.
2.9vpk*40 = 116vpk
Let's back-up for a sec. We've been working in peak voltages, and there is a good reason for that, if we figure the true capability of each stage based not only on gain but also on loadline and supply voltage. Let's convert to RMS for the sake of looking at needed input signal. 116vpk = 82vRMS.
So we need 82vRMS from our input stage, but the gain is ~44. 82v/44 = 1.86vRMS input
Somehow, your guitar needs to make more than 1.8vRMS, while still allowing for the voltage loss in the input resistor network on the input jacks. That is, if we want to comfortably drive an output tube biased with -30v, while distorting the output tube with a volume control between the tonestack and the 2nd gain stage, which is set to 7. It also assumes an average loss for the tonestack, which still may not be total worst-case.
There is no 1 answer to this stuff. You calculate, and reality-check to the degree you're willing, find the problem areas (stage gain, topology issues, needed supply voltage, etc) then go back and start the whole process over until you get what appears on paper as a workable answer. Then you build and carefully test, and find out that your calculations didn't catch everything.
The idea is to make as many maistakes as possible on paper, and figure out the answers before you spend money.
You will also find out there are standard parts available at reasonable cost, and therefore everything tends toward an average answer that's repeated in a lot of amps.