Hint. The Fourier transform preserves norms on $L^2$, i.e. $\|f\|_2 = \|\hat f\|_2$. Also $\hat{(f * g)} = \hat f \hat g$. So it is enough that you find a function which is in $L^2$ but whose square is not in $L^2$.
Edit after the comment. The proof of $\hat{(f * g)} = \hat f \hat g$ with $f,\ g \in L^2$ requires extending the Fourier transform to the space of tempered distributions. But you can actually prove what you want with its $L^1$ version and a contradiction. Assume that convolution is a Banach algebra, and take $f,\ g \in L^1 \cap L^2$. Then by assumption $\|f*g\|_2\leq \|f\|_2\|g\|_2$. From this it follows by isometry that $\|\hat f \hat g\|_2 \leq \|\hat f\|_2\|\hat g\|_2$. But since $L^1\cap L^2$ is dense in $L^2$, this must be true for all $f,\ g \in L^2$, for which you can find counter-examples. In particular, taking $g = f$, one gets $\|\hat f ^2\|_2 \leq \|\hat f\|_2^2$ i.e., $\|\hat f \|_4 \leq \|\hat f\|_2$. But a counter example can be produced for this by taking any function $f \in L^1\cap L^2$ and forming a sequence $f_n$ by scaling and normalizing it accordingly so that $ \|\hat f_n\|_2$ stays constant but $\|\hat f_n\|_4$ blows up.