This is a delayed answer, but it seems good to clarify what the OP is asking. This kind of statements: "since $f$ is concentrated in such a ball then its Fourier transform is essentially concentrated in such a ball" is quite common in certain fields, but it doesn't mean that the function and its Fourier transform are compactly supported, because this is wrong, as noticed already. People usually omit the details of what is meant, however there are several ways of formalizing this heuristic.

Fix a smooth function $\zeta$ supported in the unit cube $Q$ centered at the origin in $\mathbb{R}^n$. Then, by standard methods, we see that $|\hat{\zeta}(\xi)|\le C_N\frac{1}{(1+|\xi|^2)^{N/2}}$, where $C_N$ is a constant depending on $N$ and $\zeta$; but as $\zeta$ is fixed, we forget about it. Whatever the reason is, you want to use a cut-off function for some parallelepiped $P$, so you take the affine transformation $A$ transforming $Q$ into $P$, hence $\zeta_A(x)=\zeta(A^{-1}x)$ works as a cut-off and its Fourier transform is $\widehat{\zeta_A}(\xi)=|\det A|\hat{\zeta}(A^t\xi)$, and $|\widehat{\zeta_A}(\xi)|\le C_N|\det A|\frac{1}{(1+|A^t\xi|)^{N/2}}$ , hence we see that $|\widehat{\zeta_A}(\xi)|$ decays strongly outside the "dual parallelepiped" $A^{-t}Q$ or is "concentrated" in $A^{-t}Q$; it is in general unimportant the position of $A^{-t}Q$, but its dimensions and orientation in space. This had been basically noted by Willie Wong.

People usually replace $|\widehat{\zeta_A}|$ by $\chi_{A^{-t}Q}$ when they try to get upper bounds, because $|\widehat{\zeta_A}(\xi)|\le C_N\sum_{\nu\in A^{-t}\mathbb{Z}^n}\frac{1}{(1+|\nu|^2)^{N/2}}\chi_{A^{-t}Q}(\xi-\nu)$. As in every field, there is a toolkit you acquire after some time, so it's hard to provide a single reference of the many ways this heuristic is applied.

By the way, to say that if $f$ is supported in a cube then its Fourier transform is concentrated in the dual cube is not quite precise and depends on the context. What is true in general, is that if $f$ is supported in a cube, then $|\hat{f}|$ is "essentially" constant in translations of the dual cube.

outsidethe dual cube. (A form of the uncertainty [a.k.a. indeterminacy] principle mentioned byAlexandre Eremenko.) $\endgroup$ – Noam D. Elkies Apr 6 '17 at 3:54