Consider $\Omega = (0,2 \pi)^d$ and the negative Sobolev space $H^{-s}(\Omega)$, defined as the dual of $H^s_0(\Omega)$ for the $L^2$ inner product. Due to the simple shape of $\Omega$, we can see $H^{-s}(\Omega)$ as a Banach/Hilbert space with the inner product: $$ \langle f, g \rangle_{H^{-s}} := \sum_{n \in \mathbb{Z}^d} f_n g_n (1 + | n |^2)^{-s}, $$ where $(f_n)$ and $(g_n)$ are the Fourier coefficients of $f$ and $g$.
Question: Is $C^\infty_0(\Omega)$ dense in $H^{-s}(\Omega)$?
The answer is affirmative. In a previous version of this post I claimed the opposite, but I was wrong. (The previous version is below. Following my usual policy, I am NOT deleting it).
Claim. The space $C^\infty(\mathbb T)$ of smooth and $2\pi$ periodic functions is dense in $H^{-s}(\mathbb T)$ for all $s\in \mathbb R$.
Proof. We will prove that $C^\infty(\mathbb T)^\bot$ is reduced to $\{0\}$. So, let $g\in H^{-s}(\mathbb T)$ be such that $$ \sum_{-\infty}^\infty \frac{\hat{g}(n)\overline{\hat\phi(n)}}{(1+|n|^2)^s}=0, \qquad \forall \phi \in C^\infty(\mathbb T).$$ In particular, letting $\phi(x)=e^{imx}$, so that $\hat\phi(n)=0$ for $n\ne m$ and $\hat\phi(m)=1$, we see that $$\frac{\hat{g}(m)}{(1+|m|^2)^s}=0, $$ which implies $\hat{g}(m)=0$. Since $m$ is arbitrary, we conclude that $g=0$, as claimed. $\Box$
Remark. This is exactly the proof suggested by the OP dgontier in comments below, and also by ktoi in comments to the main question.
Remark. I find this result a bit surprising. If $f\in H^{-s}(\mathbb T)$, its Fourier coefficients $\hat{f}(n)$ are allowed to grow as $n\to \infty$. If $s>1/2$, the space $H^{-s}(\mathbb T)$ contains singular distributions such as the Dirac comb $\delta$. On the other hand, if $\phi\in C^\infty(\mathbb T)$, then its Fourier coefficients decay faster than any polynomial. So, I find it surprising that $C^\infty$ functions suffice to approximate arbitrary $H^{-s}$ elements.
PREVIOUS VERSION OF THIS ANSWER.
WARNING. There is a mistake, marked below. Now I have to go, but I am not deleting this answer, maybe someone will point out a solution, or it will be useful in some other way.
The answer is negative. To ease typing, let me assume $d=1$. Let $f_n\in C^\infty_0(0, 2\pi)$ and suppose that there is a $f\in H^{-s}$ such that $\lVert f_n-f\rVert_{H^{-s}}\to 0$. This means that $$ \sum_{k=-\infty}^\infty \frac{\lvert \hat{f}_n(k)-\hat{f}(k)\rvert^2}{1+k^{2s}}\to 0, \qquad n\to \infty.$$ In particular, for each $k_0\in\mathbb Z$, $$\tag{*} \lim_{n\to \infty} \frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} =0.$$ Now, integration by parts shows that, since $f_n$ is smooth, $\hat{f}_n(k)$ decays as $\lvert k \rvert\to 0$ (faster than any power of $k$, but we won't need this quantification). For sufficiently big $n$, we have from (*) that $$ \frac{\lvert \hat{f}_n(k_0)-\hat{f}(k_0)\rvert^2}{1+k_0^{2s}} \le 1.$$
However, not all $f\in H^{-s}$ satisfy this. For example, if $s>1/2$ then the Dirac comb is in $H^{-s}$, and $\hat{\delta}(k)=1$ for all $k$.