How to analyze the biasing of the output stage in this amplifier?
I'll assume we're calculating bias current in the same setup we'd use to measure it: the amp is idle, 0V input, 0V output (neglecting offset), no output load. All transistors are ON, linear mode (not saturated), Vbe=0.6V. I'll start by neglecting base currents.

Q6/Q7 is a current source, voltage on R9 is Q6 Vbe so current in R9 and Q7 is 9.7 mA, a common value.

Q5 is wired as a Vbe multiplier, which is a shunt regulator with feedback. There is one Vbe across R7, so Q5 Vce = Vbe/(R7/(R7+R8)) = 2V as you said. Note Q5 should be mounted on the heat sink with power transistors so its Vbe tracks Q8-9-10-11 Vbe to ensure bias does not drift (or go into thermal runaway) as the output transistors heat up.

Now there is 2V between the bases of Q8 and Q9, but we have to fit the sum of Vbe's of Q8-9-10-11 inside those 2V... and 4*Vbe is 2.4V.

If we assume Vbe=0.5V instead of 0.6V then the Vbe multiplier bias voltage shrinks in the same proportion, and it doesn't work.

This means there is a problem somewhere. Either the output transistors Vbe is quite lower than 0.6V, or the resistor values on the schematic are wrong and the Vbe multiplier actually sets a higher bias voltage than 2V.

Also the schematic introduces unnecessary crossover distortion. If you build the amp, DO NOT CONNECT the midpoint of R11-R12 to the midpoint of R13-R14. You want your drivers to run in class A push-pull so they are able to suck charge out of Q10/Q11 bases quickly when Q10/Q11 have to turn off, instead of simply relying on a 33R resistor. On the schematic as shown, when say the top transistor has to turn off, the top driver also turns off, and there is only Vbe/33R = 18mA current available to suck charges out of Q10 base. So it will take extra delay before it turns off, while Q11 is driven by Q9, and turns on much quicker. This can lead to cross-conduction and crossover distortion.