> it could be some kind of cathode follower.
No signal is taken from cathode. ipso fractal, it aint no cathode follower.
> essentially the same arrangement as a Gibson Scout. There's a cap driving it
I don't see any part of this well-copied from a Scout. You've replaced the efficient transformer with a much too small resistor, got the cathode bias 1000X bigger than the Scout's. The tube is "starved" (teeny current) and what little it squeaks out would rather go through the plate resistor than the tank.
If you can't calculate it, copy it. That includes the number of zeros on the parts. I'm not a plumber, and I got pipe-trouble. I see that plumbers use 1/4" tube for an ice-maker and 4" pipe for the sewer. There's probably a reason. So it might be unwise to extend a waste-drain with 1/4" tubing. In this case I can picture poop flowing (NOT!) in a thin hose, and stagnant water if I ran 4" pipe to an ice-maker. In electronics we can't see the poop, but we do want to know or copy "reasonable sizes".
> Would the value of the drive cap make a big difference? I thought it was just blocking DC.
If "DC block" was your ONLY goal, you'd just cut the wire.
The thing is: you want audio to get through. And the tank is a heavy load. You need a fairly big cap. Just like plumbing the toilet. While we vent so the steady ("DC") stream of air from septic to roof does not come up the toilet, we do need to pass the "interesting" stuff from toilet to septic. You need a handful of pipe or capacitor. (Can you tell how much fun I had today with poop-pipes?)
8FB3C1C's important specs are;
Type: Accutronics Type 8
Input impedance: 1,925 ohms
Output impedance: 2,575 ohms
OK, that's as good as you will get for a plate-loaded reverb driver.
The plate load resistor is a compromise between the actual load (around 2K) and the tube's plate resistance (about 7K for 12AU7 or the fat side of a 12DW7). 82 is not a compromise. 6.7K is optimum but 10K will be easier to find in the required 5 Watt size.
Note that this is 100 times bigger than you seem to have.
The cathode resistor will typically drop "a few" volts at the design cathode current. Call it 3V for starters. You are trying to drive a low-Z load with inefficient coupling, you need big current. Your little tubes run 1mA (200K plate resistor dropping about-half of 300V supply) and drive silly little pots etc, your big 6V6 suck about 40mA each and do real work (a speaker), you should be thinking like 10mA. To drop 3V at 10mA you figure 3V/10mA= 3V/0.010A= 300 ohms. Use 270 ohms. 1K may work, and save power; seems thin to me.
Note that this is 1000-3000 times smaller than your 1Meg.
The combination of plate-side 100X off and cathode-side 1000 times off suggests that driver gain and power are maybe 100,000 times smaller than they could be. Since it takes more parts than this to get BIG reverb, and you have to trim 2 triodes and a transformer to get even modest reverb, 100,000 times worse is "no" reverb.
You are starved for signal, you do NOT want the resistor dividers at the grids.
You don't want bass in the reverb (mudd) and you don't want to load your preamp and vol/tone network, so you steal the 500pFd cap and 1Meg from the nearest Fender (it works).
Bypass the cathode resistor with any handy electrolytic.
I can't make out what size tank coupler you used. Ampeg and I tend to go ~~0.5uFd, a 0.22uFd is mentioned, whichever, but something in that range. A 0.01uFd, fer example, would (in this low-Z case) only pass the tizz, not the body and voice of the guitar.