As I try to learn from advice (and not just follow it) I have a question - Abbey Road d'Enfer - you said I'm looking at 50ma of idle current - is that the right number? Earlier I added up the amperage from the tube array and it was 2.69 amps. Is that 'under load' at 2.69 and you're talking about when not under load? Just trying to get my head around what is going on here.
Also, I ordered up a slew of different value 10W resistors, I was found greatly lacking in that part of inventory. Should be here tomorrow.
I am not Abbey but I can solve your confusion.
You are confusing two important parameters of tube circuits here (I think ChatGPT is not entirely innocent).
For tubes to be operated at all, they have to be heated. This current is called heating current and it is relatively high. For example, for a 6V6 power tube you need 0.45A, i.e. 450 milliamperes. A 12AX7 preamp tube requires slightly less, at 6.3V this is 0.3A. A fat power tube such as an EL34, for example, needs 1.5A per tube, so you can see that a powerful power amplifier can already add up to a few amps (and a lot of heat).
That is one side of the coin, the heating current ensures the basic functionality of the tube.
On the other hand, there is the high-voltage operating tension, usually several hundred volts, which is often referred to as B+. Only relatively low currents flow here due to the high voltages.
Let's take your amplifier, the 6V6 will require the largest amount of current, roughly between 35mA and 50mA. This is divided between the anode current and the screen grid current when connected as a pentode, more details can be found in the relevant data sheets. The current consumption of the preamplifier tubes is almost negligible in comparison. A 12AX7 requires just over 1mA per triode system, which is almost nothing compared to the power tube.
So now you understand the mA specification from Abbey, he was talking about the B+ voltage and the total amount of current required there. This is important in your case, because this affects your "too much voltage" problem, because the series resistors that are supposed to reduce this, do it according to Ohm's law. The current of the B+ voltage is important for that, not the heating voltage!