Let $\Omega$ be an open bounded domain of $\mathbb R^d$ with smooth boundary $\Gamma$.
I am wondering if it is possible to characterise the closure of $C^{\infty}(\overline\Omega)$ with respect to the norm $$ \|u \|:=\sqrt{\|u \|^2_{L^2(\Omega)}+ \| u\|^2_{L^2(\Gamma)}}. $$ Any useful reference will be very appreciated.
Let $m$ be Lebesgue measure on $\Omega$ and $\sigma$ be surface measure on $\Gamma$, both of which may be extended trivially to measures on $\overline{\Omega}$. Then if $\mu = m + \sigma$ is their sum, your norm is just the $L^2(\overline{\Omega}, \mu)$ norm. And $C^\infty(\overline{\Omega})$ is dense in $L^2(\overline{\Omega}, \mu)$, e.g. by Stone-Weierstrass; this argument works for any measure. So the completion is (isometrically isomorphic to) $L^2(\overline{\Omega}, \mu)$, or equivalently, to the direct sum $L^2(\Omega,m) \oplus L^2(\Gamma,\sigma)$.
Previous, overcomplicated argument (don't read):
The completion is isometrically isomorphic to the Hilbert direct sum $L^2(\Omega) \oplus L^2(\Gamma)$. If you like, this is equivalent to $L^2(\overline{\Omega})$ where $\overline{\Omega}$ is equipped with the sum of Lebesgue measure on $\Omega$ plus surface measure on $\Gamma$.
Identify $C^\infty(\overline{\Omega})$ with a subspace of $L^2(\Omega) \oplus L^2(\Gamma)$ in the obvious way, by identifying $u$ with $(u|_\Omega, u|_\Gamma)$. Then your norm agrees with the norm of $L^2(\Omega) \oplus L^2(\Gamma)$, so it suffices to show that $C^\infty(\overline{\Omega})$ is dense in $L^2(\Omega) \oplus L^2(\Gamma)$.
Let $J(\Gamma) \subset C(\Gamma)$ be the "jet space" of functions $v : \Gamma \to \mathbb{R}$ which have a $C^\infty$ extension to some neighborhood of $\Gamma$. By the Stone-Weierstrass theorem, $J(\Gamma)$ is dense in $C(\Gamma)$ (with respect to the uniform norm) and therefore dense in $L^2(\Gamma)$ (with respect to the $L^2$ norm). Also, the space $C^\infty_c(\Omega)$ of compactly supported smooth functions is dense in $L^2(\Omega)$. I'll show that the closure of $C^\infty(\overline{\Omega})$ contains $C^\infty_c(\Omega) \oplus J(\Gamma)$.
Let $u \in C^\infty_c(\Omega), v \in J(\Gamma)$. Extend $v$ to a smooth function on some neighborhood of $\Gamma$ and multiply by a smooth cutoff to obtain a $C^\infty(\overline{\Omega})$ function, still called $v$, whose support is disjoint from that of $u$. Now consider a sequence of smooth cutoff functions $\phi_n : \overline{\Omega} \to [0,1]$ which equal 1 on $\Gamma$ and converge pointwise to 0 on $\Omega$. The functions $u_n = u + \phi_n v$ are in $C^\infty(\overline{\Omega})$. We have $u_n|_\Omega \to u$ pointwise and boundedly, hence in $L^2(\Omega)$, and $u_n|_{\Gamma} = v$ for all $n$. Hence $u_n \to (u,v)$ in $L^2(\Omega) \oplus L^2(\Gamma)$.