Simple power amplifier biasing calculations
First, please put proper DC supplies on your simulation.

Since the push/pull output stage has a unity gain, the DC level on Q3's collector, \$V_{C_{Q3}}\$, will be seen on VOUT node as \$V_{OUT}=V_{C_{Q3}}+V_{BE_{Q2}}\$ -neglecting the drop on 0R1 resistor.

In order to set VOUT node to the midpoint of ±12V, which is 0VDC, your first aim should be setting \$V_{C_{Q3}}\$ to about -0.6VDC. 

Also, the common emitter amplifier formed by Q3-R1-R2-R5-R6 has a gain of R1/R5 and equals the total gain of the whole amplifier circuit.


Quiescent current for biasing diodes can be anything between 2-10mA. Let's select 5mA.
Gain is \$K_v=R1/R5\$ and input voltage is 70mVrms = 0.1Vp. Since the peak value of output signal can be VCC=12V before clipping, \$K_{v(max)}= 12 / 0.1=120\$ for unclipped maximum swing --not necessary to set that high though.


Example:

Let's set the gain to 2.







simulate this circuit &ndash; Schematic created using CircuitLab

So \$R1/R5=2\$. In order to have -0.6VDC on collector of Q3, voltage drop on R1 should be \$V_{R1}=(+12V - (-0.6V) - 1.2V)=11.4VDC\$. Let's select \$I_{bias}=5mA\$, so \$R1=11.4V/5mA=2.2k\$ so R5 will be 1k1. Since the voltage drop on R5 will be \$V_{R5}=1.1k \cdot 5mA = 5.5V\$, the voltage on the base node of Q3 should be \$V_B=-12 + (5.5 + 0.6) = -5.9V\$. For this purpose, I chose R2=33k and R6=12k.

According to simulation results, quiescent output DC level is 0.3VDC which is tolerable. If you set R1 = 5k1 and R5 = 2k5 then the quiescent DC output on VOUT node will be less than 0.1VDC, but the current passing through the biasing diodes will be reduced to about 2.5mA which is fairly enough.

If we select \$K_v=10\$ and \$I_{bias}=2mA\$ then we can find that (with the calculations above) R1=5k6, R5=560R, R2=33k and R6=2k7.

You can tune the circuit according to your gain needs.