No it is not true.
First note that we can improve pointwise convergence to uniform convergence of $f_{n} \ast g_{m} \to f \ast g$. This is an easy consequence of Hölder's inequality (markup doesn't like my subscripts, so I'll use $\|\cdot\|$ for the $L^{2}$-norm): \[ |(f \ast g)(x)| \leq \int |f(x - y) g(y)|\,dy \leq \|f\| \|g\|. \] As $f_{n} \ast g_{m} - f \ast g = f_{n} \ast (g_{m} - g) + (f_{n} - f) \ast g$ we get \[ |(f_{n} \ast g_{m})(x) - (f \ast g)(x)| \leq \|f_{n}\| \|g_{m} - g\| + \|f_{n} - f\| \|g\|. \] Since $\|f_{n}\| \to \|f\|$ and both $\|f_{n} - f\| \to 0$ and $\|g_{m} - g\| \to 0$, the right hand side can be made arbitrarily small independently of $x$. It follows in particular that $f \ast g$ is continuous and vanishes at infinity (that's one reason why I prefer writing $C_{c}$ for functions with compact support and $C_{0}$ for functions vanishing at infinity).
But we cannot do better, i.e., get $L^1$-convergence without further hypotheses. To see what happens, I prefer to ignore the condition that $f_{n} \in C^{\infty}$ but rather look at the function $f_{n} = \frac{1}{n}[-n,n]$. Then $f_{n} \to 0$ in $L^{2}$ but not in $L^{1}$. By the above argument we have $f_{n} \ast f_{n} \to 0$ uniformly on $\mathbb{R}$. On the other hand it is easy to see that for $x \in [-\frac{n}{2},\frac{n}{2}]$ \[ (f_{n} \ast f_{n})(x) = \int f_{n}(y) f_{n}(x-y)\,dy \geq \frac{1}{n^{2}} \frac{n}{2} = \frac{1}{2n} \] (the intervals $[-n,n]$ and $[-n-x,n-x]$ have an overlap of length at least $\frac{n}{2}$) and therefore we have the estimate $\int (f_{n} \ast f_{n}) \geq \frac{1}{2}$. Thus we cannot have $\int f_{n} \ast f_{n} \to \int f \ast f = 0$. It is now easy to cook up an example along these lines with smooth functions.