$\Omega\subseteq \mathbb{R}^n$ is a non-empty, bounded, connected open set with smooth boundary here. Edit: for clarity, by $C^{\infty}_0(\overline{\Omega})$ I mean the smooth functions which vanish on $\partial \Omega$.
Take $n=1$. It's easy to find elementary functions $f_n\in C^{\infty}(\mathbb{R})$ so that $(\|f_n'\|_{L^2})/(\|f_n\|_{L^2})\to \infty$. Is there a simple choice of even elementary functions $f_n\in C^{\infty}_0([-1,1])$ with the same property? If so, then I suppose one could use polar coordinates $F_n(r,\vec{\theta})=f_n(r)$ to generalize to $\Omega = B_n(0,1)\subseteq \mathbb{R}^n$, and then use a translation with dilation transformation to generalize to arbitrary open balls, and hence arbitrary $\Omega$. Is that correct? Under the assumption it is, it suffices to find the required functions for the case $\Omega = (-1,1)$.
I was thinking of modifying the standard bump function $$f_n(x)=\Phi_n(x) = \begin{cases} e^{-1/(1-x^{2n})} & \text{if } x \in (-1,1)\\ 0 & \text{otherwise} \end{cases};$$ numerical calculations indicate $\|f_n\|^2_{L^2}\to 2e^{-2}$ and $\|f'_n\|^2_{L^2}/\|f_n\|^2_{L^2}-n\to 0$, so I'd be done if I proved both claims. I think I've worked out the almost uniform convergence $f_n\to e^{-1}\chi_{[-1,1]}$ to prove the first part, but I'm not quite sure about the latter. It is equivalent to showing $$\int_{-1}^1 2m\frac{x^{2(m-1)}}{(1-x^m)^4}e^{-2/(1-x^m)}dx \to 2e^{-2} = \int_{-1}^1 e^{-2} dx$$ as $n\to \infty$, where $m=2n$. Is there an easy way to prove this? I'm not sure if we again have almost uniform convergence. If you are certain proving the claim would be messy, can you come up with a simpler choice of $f_n$? (I would like them to be elementary functions.)
I know most of the question is about your sequence, but do you mind if I give you a slightly different sequence to prove the statement in the title?
In $\mathbb R^d$, without loss of generality, assume $B_2(0)\subseteq\Omega$ (true up to translations and dilations). Consider $f(x)=\Phi_1(x)$ as in your definition (by taking $|x|$ instead of $x$, so it is well-defined in all dimensions). Define $$f_k(x)=f(kx).$$ From the definiiton of $\Phi$, it is clear that the support of $f_k$ equals $\overline B_{1/k}(0)$, so $f_k\in C^\infty_0(\Omega)$ for all $k$. By explicit calculations (using the change of variables $x\mapsto y=kx$), it holds $$ \|f_k\|_{L^2}=k^{-d/2}\|f\|_{L^2}, $$ $$ \|\partial_{x_j}f_k\|_{L^2}=k^{-d/2+1}\|\partial_{x_j}f\|_{L^2}. $$ So, the ratio $\|\partial_{x_j}f_k\|_{L^2}/\|f_k\|_{L^2}$ diverges as $k$ goes to infinity, so the operator cannot be bounded. This proof works uniformly well on all $\mathbb R^d$, $d\geq 1$.
This is a standard scaling argument that comes useful in many interesting settings. The core idea is that when you dilate functions by a factor of $k$, different Lebesgue or Sobolev norms scale as a different power of $k$.