The answer is yes, the discrete Fourier transform (DFT), of which the fast Fourier transform (FFT) is an efficient implementation, can be interpreted quite literally as a filter bank. Recall the formula for the DFT:
$$
X[k] = \sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi k n}{N}}
$$
The "filterbank" interpretation for the above equation goes something like this:
In order to get the $k$-th DFT output $X[k]$, we start with the input signal $x[n]$.
Multiply $x[n]$ by a complex exponential function $e^{\frac{-j2\pi k n}{N}}$. By the frequency-shifting property of the DFT, this has the effect of shifting the content at frequency $\frac{2\pi k}{N}$ down to zero frequency.
Convolve the result with a filter that has the impulse response:
$$
h[n] = \begin{cases}
1, && 0 \le n \le N-1 \\
0, && \text{otherwise}
\end{cases}
$$
This filter (known as a moving average or a "boxcar") has a lowpass characteristic.
Thus, we can view the DFT as placing a bank of uniformly-spaced critically-sampled filters across the signal's spectrum. Each filter has the same frequency response as an $N$-sample moving average filter, centered at frequencies $e^{\frac{j2\pi k n}{N}}, k = 0, \ldots, N-1$. I described this similarly in an answer to a previous question here.