Let $\mu$ be a non-zero finite Borel measure with compact support on $\mathbb{R}^d$.
Facts
If $\widehat{\mu} \in L^2$, then $\mu$ is absolutely continuous with respect to Lebesgue measure with an $L^2$ density.
If $\widehat{\mu} \in L^1$, then $\mu$ is absolutely continuous with respect to Lebesgue measure with an $L^{\infty}$ (in fact, continuous) density.
If $\widehat{\mu} \in L^p$, for some $1 < p < 2$, then $\mu$ is absolutely continuous with respect to Lebesgue measure with an $L^{p'}$ density, $p'=p/(p-1)$. (The proof is basically the same as the $L^1$ case; just using Hausdorff-Young at the end).
Question
If $\widehat{\mu} \in L^p$, for some $2 < p \leq \infty$, then what can be said about $\mu$?
The knowledge that $\hat \mu \in L^\infty$ adds nothing new, since the Fourier transform of any finite measure is in $L^\infty$.
If $\hat \mu\in L^p$ with $2<p<\infty$, then $\mu$ has no atoms, i.e., $\mu(\{x\})=0$ for every $x$. See this MathOverflow post.
Further along these lines, being in $L^p$ with not very large $p$ precludes $\mu$ from being supported on a lower-dimensional surface. Specifically, if $\hat \mu\in L^{2d/k}$, then $\mu$ cannot be supported on a $k$-dimensional $C^1$-smooth surface. This is a theorem of Agranovsky and Narayanan, quoted in this thesis which also has other references to literature on the topic.