I obtained some partial results for my question. 

 - By the [Hausdorff-Young inequality][1], $\|\widehat{1_A}\|_q\leq \|1_A\|_p=1$ 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. 

 


  [1]: http://en.wikipedia.org/wiki/Riesz%E2%80%93Thorin_theorem#Hausdorff.E2.88.92Young_inequality