Class AB amplifier Design
For basing diodes, 1-5mA range is quite acceptable. 

But does not finish with this. You need some other info:

As you might know, output current passes through collectors. So each output transistor's base will need a current of \$I_B = I_C / \beta \$. Thus, R1 and R2 should also allow enough base current for output transistors.


Determine output power (\$P_O\$) and calculate output voltage (\$V_O\$) and current (\$I_{O}\$). Also, this circuit has a unity voltage gain:


$$V_{O-rms} = \sqrt{P_O \cdot 4\Omega} \ ,\ \ \ \ V_{O-pk} = V_{o-rms} \cdot 1.41$$
$$I_{O-rms} = \sqrt{P_O / 4\Omega} \ ,\ \ \ \ I_{O-pk} = I_{o-rms} \cdot 1.41$$

Select proper transistors having enough \$V_{CE}\$ for \$V_{O-pk}\$ and \$I_C\$ for \$I_{O-pk}\$. Select Vcc for enough voltage swing: \$V_{CC} = 2\cdot V_{o-pk} + 1V\$


From datasheets, calculate required base currents from \$I_B = I_C / \beta\$. Note that \$\beta\$ value can be quite low (e.g. between 10 and 25) for high output currents.
Calulate R1 and R4 for diode bias current and transistor base current.


Input coupling cap is calculated from minimum input frequency (\$f_L\$). If we assume input impedance is parallel equivalent of R1 and R2 (say \$R_i = R1 || R2\$) then input coupling cap is \$Ci = 1/(2\pi f_L R_i)\$. If you don't have any info, put a 10uF electrolytic and test. 

Output coupling cap is also calculated from minimum input frequency: \$C_o = 1/(2\pi f_L \cdot 4\Omega)\$. If you don't have any info, put a 2200uF electrolytic (+ to common emitter, - to load resistance) and test.