Let $M \subset L^2(\mathbb{R})$ the space of rapidly decreasing functions, and $\mathcal{F}_0 = \mathcal{F}\lvert_M$ the restriction of the Fourier transform to $M$. Then $\mathcal{F}_0\colon M \to L^2(\mathbb{R})$ is a linear isometry, and since $L^2(\mathbb{R})$ is complete, it can be extended to a linear isometry $\Phi \colon L^2(\mathbb{R}) \to L^2(\mathbb{R})$.
Now let $f \in L^1 \cap L^2$. Then there is a sequence $(s_n)$ of rapidly decreasing functions (we can even choose smooth functions with compact support) that converges to $f$ simultaneously in the $\lVert\,\cdot\,\rVert_1$ and $\lVert\,\cdot\,\rVert_2$ norms. (If that's not known, it follows relatively easily by approximating $f$ first with a bounded function with compact support simultaneously in both norms, then Hölder's inequality with the characteristic function of the compact support yields the result.)
Then $\mathcal{F}_0(s_n)$ converges uniformly to $\mathcal{F}(f)$, since $\lVert s_n - f\rVert_1 \to 0$, and $\mathcal{F}_0(s_n)$ converges to $\Phi(f)$ in $L^2(\mathbb{R})$. All sequences that converge in $L^2$ contain subsequences that converge pointwise almost everywhere to the limit, so that means $\Phi(f) = \mathcal{F}(f)$. Hence, since $\Phi$ is an isometry,
$$\lVert \mathcal{F}(f)\rVert_2 = \lVert \Phi(f)\rVert_2 = \lVert f\rVert_2.$$