We have trace operator which allows us to define boundary values of an $H^1$ function. This is because of the fact that $C^\infty$ is dense in $H^1$ under the $H^1$ norm, I believe.
I'm sure either $C^0$ or $C^\infty$ is also dense in $L^2$ in the $L^2$ norm, so why no trace operator in this case? Or am I wrong?
The problem is that even though you can of course define a trace $T: C^\infty(\overline \Omega) \to L^2(\partial \Omega)$, to be able to extend $T$ to all of $L^p(\Omega)$ in a meaningful way, it is not sufficient to have any old operator $T$, but you really want $T$ to be continuous, i.e. there would need to be a constant $C >0$ such that
$$\Vert Tf \Vert_{L^2(\partial \Omega)}\le C \Vert f \Vert_{L^2(\Omega)}$$
In this case we would be able extend $T$ to an operator $T: L^2( \Omega) \to L^2(\partial \Omega)$ nicely. This is the case if you take the $H^1$ norm instead of the $L^2$ norm. However, it's a good exercise to show that such a $C$ does not exist for $L^2$ (or indeed for any $L^p$ space). Consider something simple like $\overline \Omega = [0,1]$.