Making the Fourier transform quantitative

There are various mathematical formulations of this phenomenon and some of them are quite quantitative.

First, there is the classical Heisenberg uncertainty principle which roughly states that the product of the variances of a functions and its Fourier transform is bounded from below, independently of the function.

Then there is the uncertainty principle of Donoho and Stark. Its discrete form says that for every $x\in\mathbb{C}^N$ it holds that product of the number of non-zero entries in $x$ and the number of non-zero entries in its Fourier transform $\hat x$ is greater than $N$. The continuous version is a bit more complicated: Call some function $f$ $\epsilon$-concentrated on a set $T$ if there is some $g$ with support in $T$ such that $\|f-g\|_2\leq \epsilon$ (i.e. the $L^2$-norm of $f$ outside of $T$ is smaller than $\epsilon$). The uncertainty principle says that for $f$ and $\hat f$ with unit norm such that $f$ is $\epsilon_T$ concentrated on some set $T$ and $\hat f$ is $\epsilon_W$ concentrated on some set $W$ it holds that $$|T||W|\geq (1- \epsilon_T-\epsilon_W)^2.$$

There is much more to say, e.g. there is Lieb's uncertainty principle for the short-time-Fourier transform, there are results of "time-and-band-limiting operators" by Landau/Pollack and Slepian/Landau. I can recommend the books "Foundations of Time-Frequency Analysis" by Karlheiz Gröchenig and "Ten Lectures on Wavelets" by Ingrid Daubechies for further reading.

Notice that $\|f1_{B(0,\delta)}\|_{L^2}\to 0$ as $\delta\to 0$. So the answer to the question is no because it should reasonably include the requirement $\varepsilon'<1$. -- Rapid decrease of the Fourier transform $F(f)$ implies smoothness of $f$ but, without further assumptions, it does not imply localization.