If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\|g\|_{L^1}$.
Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof. (Except for special case: if $g$ is in $L^1$, then we can chop it into pieces with disjoint supports and small $L^1$ norm, and then use $f$ to move the Fourier transforms of pieces far from one another.)
If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\|g\|_{L^1}$.
Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof.