Quantifying goodness of amplitude estimation
Your understanding of the FT as several extremely narrow bandpass filters is fine. However, you mix up the terms SNR and noise variance. SNR is defined by

$$SNR=\frac{E(s(t)^2)}{E(n(t)^2)}$$

where $E[n(t)^2]=\int_\mathbb{R}S(f)df$ due to the Wiener-Khintschin Theorem ($S(f)$ is the PSD of the noise). So, if you have a measured signal $x(t)=s(t)+n(t)$ with $s(t)=A\sin(2\pi f_0t)$ and $n(t)$ is bandlimited AWGN with bandwidth $B$ and density $N_0$, the SNR decreases when the noise bandwidth increases. So, in order to have a good SNR, you should keep the bandwidth as small as possible (as narrow, as your $s(t)$ permits). 

However, your problem is to estimate the amplitude $A$ of $s(t)=A\sin(2\pi f_0t)$ where $f_0$ is known to you. Optimally, you take the Fourier transform of your signal and see what is the value of $X(f)$ at the frequency $f_0$. What will you see there? Does this measurement depend on SNR? Does this measurement actually depend on the noise bandwidth? No, the measurement $X(f_0)$ is independent of the noise bandwidth (as long as the noise has contribution at $f_0$). So, by performing bandpass filtering, increasing/decreasing bandwidth you will not get any better estimates.

The only thing which you can do is to increase the duration of the measurement window. If you have a finite length measurement of duration $T$, your frequency domain essentially has resolution $1/T$. Hence, on each frequency bin (assuming the DFT) the noise variance is $N_0/T$ (assuming AWGN). So, the longer the measurement, the better the estimate. 

You can also understand this from a different point: What you are interested in, is the value 

$$\rho=\frac{\int_{f_0-\Delta_f}^{f_0+\Delta_f}S(f)df}{\int_{f_0-\Delta_f}^{f_0+\Delta_f}N(f)df}$$

where $2\Delta_f$ describes the frequency domain resolution, $S(f)$ is the signal spectrum and $N(f)$ is the noise spectrum. The numerator is always the same (independent of $\Delta_f$) since $S(f)$ is just a (weighted) Dirac at $f_0$. But, the denominator depends on $\Delta_f$.