The inequality \begin{equation*} \|\widehat{f}\|_{L^\infty(\mathbb{R}^d)} \le \|f\|_{L^1(\mathbb{R}^d)} \end{equation*} is well-known. Moreover, as far as I understand, it has no `inverse version' with exchanged indices (not even up to a multiplicative constant). A counterexample is the function $x\mapsto\sin(x)/x$ that does not belong to $L^1(\mathbb{R})$, but whose Fourier transform (with the proper choice of $2\pi$'s) is a characteristic function, and thus bounded by 1.
I was wondering if working on the torus $\mathbb{T}$ would help in order to have an inverse inequality. Namely, if it is true that \begin{equation*} \|f\|_{L^1(\mathbb{T})} \le C \|\widehat{f}\|_{\ell^{\infty}}. \end{equation*} The counterexample for $L^1(\mathbb{R})$ above works because of a lack of integrability at infinity, which is not an issue on $\mathbb{T}$. I still definitely suspect that the inequality I wrote is not true, but I failed at finding a counterexample.