Effective impedance calculation for noise removal from supply
First of all your confusion stems from thinking about R1, R2, and R3 as if searching for thevenin values, when they are independent of each other. R1 with R2 divide Vcc in half, and if it is known you could say they divide the noise on the 5 volt line in half. R3 should be two or more times the value of R1 so the signal input impedance is high enough not to load down most signal sources.

Your time constant for noise is just T = \$ 1.1 * R1 * C2\$. The pass frequency is 1/T = 10 HZ. You are thinking minimal values when many engineers think of the maximum value possible in terms of both cost and space allowed. Your 1.6 uF capacitor would become 4.7 uF (5 HZ) or even 10 uF (1 HZ). Remember that residual noise at C2 affects the DC offset at the op-amp + input, which is amplified by the op-amp gain setting.

You want to go beyond just filtering out noise that is audible, you want to filter out  minor low frequency 'thumps' in voltage due to other loads on the 5 volt line. Also you would be staying with know off-the-shelf capacitor values and the closest to 1.6 uF is 2.2 uF.

Analog Devices states this is a more stable single-supply design as R3 isolates C2 from the + input, allowing a stable Vref to be created where R1, R2, R3 and C2 meet. R3 taps into this stable and quiet DC source as bias for the op-amp equal to 1/2 Vcc, or 2.5 volts in this case. R3 also sets the audio input impedance for this amplifier. To avoid noise and hum pickup I would keep R3 under 100 K.

As the value of C2 goes up it acts as a reservoir for stable DC bias current. It is the bias for the op amp so it MUST be stable with essentially zero noise. Combined with adequate power supply decoupling there should be no self induced noise in the op-amps output. In terms of board space and cost a 100 uF for C2 (0.1 HZ) would be optimal for noise roll-off. The R1, R2 and R3 thevenin only has meaning if C2 is not there to dominate as a low-pass filter.