So, I have been thinking about this question, but I cannot find an answer. What I think is the following: If $f\in L^2(\mathbb R)$ is, e.g., VMO, then it is "close" to being continuous (as VMO is the closure of uniformly continuous functions in the BMO norm). So, $\hat f$ should be close to being $L^1$. But I have no clue what this "close" might be. Maybe weak $L^1$?
Any ideas?