calculating sinewave frequency for a low pass filter with only cutoff frequency, input sinewave amplitude and output sinewave amplitude?
Assumptions:
There are many traditional and fundamental ways to go about this in which you could use various filter equations and sampling techniques to fine tune a result that would give you the wide band operation of your filter. I propose the below solution with the assumption of a relatively ideal low pas filter and simple hand-calculated method given that your filter behaves as a simple low-pass filter. In reality you would use much more complicated filter assumptions to estimate the filter as mentioned in other comments.

Background

We essentially need to solve the low pass filter response, as @Bart mentioned.

$$H(s) = {{1} \over 1 + (sT)} $$

Where s is a frequency Laplace variable and T is the time constant.
In this example it is enough to just do magnitude of this equation. We can also replace s with jw in this case so that we can look into the specific frequency response.

$$|H(jw)| = {{1} \over \sqrt{1 + (wT)^2}} $$

Knowns:
$$Ain, Ao, fc$$

Where Ao is the output amplitude you know, Ain is the input amplitude you know and fc is the cut off frequency.

Solution:

First we find T using the cut off frequency. We know that at cut off the amplitude is -3db. To get dB from the |H(jw)| we must square it and take the log and then multiply by 10.

$$10\log{|H(jw)|^2} = 10 \log{ {{1} \over \sqrt{1 + (wT)^2}^2} \ }$$
$$-3 = {10 \log{1} - 10 \log{( 1 + (wT)^2 )} \ }$$
$$-3 = - 10*\log{(1 + (wT)^2)}$$
$$0.3 =\log{(1 + (wT)^2)}$$

$$1 + (wT)^2 = 2$$

$$T = \sqrt{{1} \over 2 \pi fc} $$ 

Now that you have T, simply plug it back into your magnitude equation and solve for fc.

$$|H(jw)| = {{1} \over \sqrt{1 + (2 \pi fc T)^2}} $$
$${{Ao} \over {Ain}} = {{1} \over \sqrt{1 + (2 \pi fc T)^2}} $$

I'm not going to go through this part with variables since actual numbers will make it a lot faster, but I hope you get the picture.