Monitor speaker volume from input lines
You want to monitor a &plusmn;2.5 Vpp audio signal with a 0-3.3 V A/D.  This means you need to reduce the amplitude and shift the result.  The gain must be 3.3 / 5 = .66, or preferably a little less.  You say this signal is being used to drive a speaker, so should be quite low impedance.

A capacitor, as you show, makes sense so that you can completely decouple the input and output DC levels.  However, you don't need 4 resistors, and they don't need to be such high values.  This should do:



R1 and the parallel combination of R2 and R3 form a voltage divider.  R2 // R3 = 5 k&Omega;.  The gain of the divider is therefore (5 k&Omega;)/(7.7 k&Omega;) = 0.65.  5.0 V in results in 3.25 V out.

C1 decouples the DC levels between IN and OUT.  It is working against 7.7 k&Omega; impedance.  At 2 &micro;F, the rolloff frequency is 10.3 Hz, so 20 Hz and up is solidly in the pass band.

The values of R2 and R3 should really be dictated by the input impedance requirement of the A/D you are using, which you haven't told us.  All three resistors are in parallel for the purpose of determining the impedance OUT is being driven with.  In this example, that is 1.8 k&Omega;.  You can scale all the resistors by the same value and still get the same attenuation and DC bias.  However, the impedance driving OUT is scaled by the same value.  For example, if you really wanted 1 k&Omega; output impedance, then scale all resistors by about 0.57.  If you lower the resistances, you have to raise the capacitance to maintain the same high pass filter rolloff frequency.

I have no idea what you mean by audio being "deteriorated" by this circuit.

Added

Yes, when calculating the output impedance, you assume the input is being driven with zero impedance.  Put another way, we are looking at the result of some voltage on IN.  What factors cause that particular voltage don't matter.  The voltage is what it is.  Given that voltage, what will be the voltage on OUT, and how will that change as OUT is loaded?