I obtained some partial results for my question.

• By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=m(A)^{\frac{1}{p}}$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
• The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.

EDIT. I found papers in which the problem of decay and integrability of $\widehat{1_A}$ has been studied and largely answered, at least for domains with some regularity.

This recent paper by Lebedev proves the following:

If $A\subset \mathbb{R}^n (n\geq 2$) has $C^2$-boundary (and finite measure), then the exponent $p=\frac{2n}{n+1}$ is ''critical'' for the integrability of the Fourier transform $\widehat{1_A}$, that is, $\widehat{1_A}\in L^p$ if and only if $p>\frac{2n}{n+1}$. Actually, even the case where $A$ is a ball shows that this cannot be improved. The same conclusion was proved by Herz if $A$ is convex but no regualrity is assumed.

A bit surprisingly, if $\partial A$ has less smoothness, namely it is just $C^1$, then for $n=2$ it is possible that $\widehat{1_A}\in L^p$ for $p>1$.

Also the decay of $\widehat{1_A}$ has been studied; for example in This paper of Svensson and This paper of Brandolini.

According to the first paper, for bounded sets $A$ that are closures of open sets with $C^{\infty}$-boundary, we have $\widehat{1_A}(\xi)=O(|\xi|^{-\frac{n+2}{2}})$ if and only if the Gaussian curvature of $A$ is nonzero everywhere.

Despite these results, it might be that for general measurable sets we cannot say anything; I have not found references yet.

I obtained some partial results for my question.

• By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=m(A)^{\frac{1}{p}}$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
• The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.

I obtained some partial results for my question.

• By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=1$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
• The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.

I obtained some partial results for my question.

• By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=m(A)^{\frac{1}{p}}$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
• The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.