Can you just stop? This is plainly wrong. Can you draw a suitable equivalent circuit with noise voltage sources and noiseless resistance and analysis the circuit? If you don't know how to do it then it's no help from me here. I remember some Texas instrument opamps have datasheet with how to calculate noise. You may wanna check those out.I was hoping for answers from those here more knowledgeable than I am. But I also studied this on my own to see if I could find the answer. What I read is that thermal noise doesn't depend on the voltage across or current through the resistor -- other than the indirect effect that higher voltage draws more current which heats it up, which increases thermal noise. This implies that thermal noise is inherently voltage/power. The dB ratio for thermal noise is given relative to an arbitrary 1 V reference. But that is arbitrary. Since it is the voltage & power of thermal noise that is constant depending on temperature but not on voltage, the actual dB ratio of noise depends on the voltage across the resistor.
For example, that 9400 Ohm resistor mentioned above has 1.74 uV of voltage noise at room temperature. If you pass a 1 V signal across it, that noise is 115 dB below the signal. But if you pass a 50 mV signal across it, that noise voltage doesn't change so it's only 89 dB below the signal.
The practical audio take-away is that there's no free lunch when attenuating signals to low levels -- thermal noise becomes a bigger factor for small signals.
If assuming the output stage doesn't have noise, and you turn the pot all the way down the noise is zero, period. If you still don't understand, I suggest learning KVL to analyze the circuit.