It's really hard to tell from your description. A good description of the noise and when it happens would be a good start.
You already said that the noise happens when you play a note. What happens if the volume is turned down? Either using the control on the electronic piano or by using the pot in your circuit. If the noise goes away then you can assume that the problem is either with the signal level on the input, or that you're asking the amp to put out more than the rated 3.5 watts.
If I were a betting man, I would guess this is the case. You're using a 3.5 watt amp to power a 15 watt speaker. Normally for a 15 watt speaker you would be using a 15 to 30 watt amp. So your amp is undersized.
Adding a cap like what @Manmanguruman suggests isn't a bad idea, but I would hope that there is already some caps inside the amp module. Still, you would use the largest cap that is somewhat practical. Start with about 470 uF and higher.
The pot should have a log taper, not a linear taper. Changing the pot will make the volume control more useful-- but will do nothing for your noise problem.
It is also possible that you have a signal loading problem with the piano. Basically, the piano might not be able to handle having an amp plugged into it. I would hope the piano was designed better, but I have seen some products do some stupid things. Turning the volume down with your linear/log pot should help this issue.
Other than that, the only thing I can think of would be a component failure. Something like the speaker, amp, battery, or piano actually being broken in some way.
First thing I notices is the time constant with your input capacitance and 1Mohm resistor - it's 47 seconds or equivalent to a high pass 3dB frequency of 0.0034 Hz.
This might take ages for your inputs to settle - make the input capacitor 47nF is my recommendation - this puts the high pass frequency at about 3.4Hz. Even a 10nF wouldn't be unreasonable (high pass of 16 Hz).
Your op-amp is a TL084 so I'm less worried about input current leakages causing offset voltage problems AND I'm happier with non-polarized caps at the inputs because if it is an electret mic and has a bias voltage on it, the electrolytic (have to be polarized correctly) input caps will become reverse biased and this may indeed be a possible reason why you see a problem.
Regarding your 50k pots, these will potentially give higher noise than say 5kpots and with 5k pots you can reduce R5, R10, R15 and R16 to something like 10ks also reducing noise. Noise is proportional to the square root of resistance so always best to go lower a bit if you can.
C7 and C12 are also problematic in the the signal they see may become reverse polarized. The time constants you are using around these circuits are very large given that those caps are 47uF - go for 1uF and non-polarized. U2 just appears to be a gain of ten stage so lose the centre biasing - 0V is centre and that is just fine assuming your Vcc and Vdd are largely equal and opposite.
You are using TL084 output amplifiers but this is really daft because you have an output pot that will likely introduce an impedance of about 25k (mid point). Just use pots and get rid of the extra 3 op-amps that feed the individual outputs. C9, C10 and C11 are not needed either or if you feel happier with them use non-polarized 1uF caps.
Best Answer
If the voltage is increased by G then the signal has increased by: -
\$20\cdot log_{10}(G)\$ decibels.
This would also mean the same decibel increase in SPL. However, for "A" weighting you should really take account of the actual SPL level before making comparisons. A quote from wiki says this: -
The reason is due to non-linearities in the ear and is typicfied by the above-mentioned fletcher Munson curve: -
But history has overtaken the FM curves and the equal loudness contours (shown in red above) are a better fit for what the ear does.
So, if your question was more aligned to understanding loudness as perceived by the ear, you must take into account the actual sound level and apply the more appropriate equal loudness curve.
EDIT to service extra question and comment
If a signal voltage rises by 6 dB then the power also increases by 6 dB and the sound pressure level also increases by 6 dB. It so happens that to increase the voltage by 6 dB you must double it and if the power increases by 6 dB it is quadrupled. This is because power is proportional to voltage squared hence a doubling of voltage (or pressure) results in four times the power.
So we often say a doubling of power is 3 dB and if it is doubled twice (i.e. made 4x bigger) it increases by 6 dB. Note that 6 dB is a slight approximation to the real figue i.e.: -
\$20\cdot log_{10}(2)\$ = more closely 6.0205 dB and
\$10\cdot log_{10}(4)\$ = the same 6.0205 dB
This is because \$log_{10}(x^2)\$ = \$2\cdot log_{10}(x)\$