Common Emitter Coupling
I think you understand the theory but have missed something simple.


As far as the AC signal \$ V_{IN} \$ is concerned R1 and R2 are in parallel as the ends away from C1 are at fixed potential. So, work that out and call it \$ R_{||} \$ as you have done.
C1 is the input decoupling capacitor. It prevents the bias from R1 and R2 being leaked away or shunted out through the input. 
The bigger you make C1 the lower the cut-off frequency becomes and you will get extended bass response in an audio circuit. (C1, R1 and R2 form a high-pass filter.) For cost, size and because we're engineers we choose one that's appropriate for the application. 
Decide your cut-off frequency and calculate the required value using your formula 


$$ C1 = \frac {1}{2 \pi f_{LOW} R_{||}} $$ 


  The capacitance shouldn't be to high or too low because it also puts on a high/low-pass filter on the input of the transistor.


It's a high-pass filter only. The sanity check on this is to take the extreme low-frequency test case - DC. It obviously blocks DC so it's high-pass. (You can use this technique to do a quick circuit analysis in lots of applications.)