Does there exist a sequence of compact operators (not necessarily linear) $T_n: H^1(\mathbb{R}^N)\to L^2(\mathbb{R}^N)$ such that, for every $u\in H^1(\mathbb{R}^N)$, $$ \lim_{n\to\infty} n\|T_n(u) - u\|_{L^2(\mathbb{R}^N)}=0? $$
For example, let us denote by $\chi_n$ the characteristic function of the ball $B(0,n)\subset\mathbb{R}^N$. Then, the operators $T_n:H^1(\mathbb{R}^N)\to L^2(\mathbb{R}^N)$ given by the formula $$ T_n(u) = \chi_n u $$ are compact (in view of Rellich–Kondrachov theorem) and, for every $u\in H^1(\mathbb{R}^N)$, $$ \lim_{n\to\infty}\|T_n (u) - u \|_{L^2(\mathbb{R}^N)}=0, $$ but I know nothing about the rate of convergence.
Recall that the operator $F$ from a Banach space $X$ to another Banach space $Y$ is compact, if the image under $F$ of any bounded subset of $X$ is a relatively compact subset of $Y$.
Ok, I found a partial answer. If we demand that $T_n$ is a linear compact operator then the construction is impossible. Indeed, assume there is a sequence of compact linear operators $(T_n)$ such that for every $u\in H^1(\mathbb{R}^N)$ we have $$ \lim_{n\to infty} n\|T_n u - u \| = 0. $$ Let us denote $L_n = n(T_n - I)$, where $I$ is the identity. By the Banach-Steinhaus theorem (uniform boundedness principle), since $L_n$ is pointwise bounded, there is $M> 0$ $$ \forall n\qquad \|L_n\| \leq M, $$ in the operator norm. Therefore, we have $$ \|T_n- I \|\leq \frac{M}{n}\iff T_n \to I\text{ in }\mathcal{L}(H^1,L^2) $$ a contradiction, since $I$ is not a compact operator.