Will this sequence of polynomials converge to a Hermite polynomial pointwise?

Yes, we do indeed have $H_{m,n}(x)\to He_m(x)$. More generally, for any monic degree $m$ polynomial $p$, setting $p_n=L^np$, then $n^{-m/2}p_n(\sqrt nx-n)\to He_m(x)$ as $n\to\infty$. As the probabilists' Hermite polynomial has distinct real roots, this implies that the zeros of $p_n(\sqrt nx-n)$ tend those of $He_m$.

The idea is to write $$ p_{n+1}(x)=p_n(x)+p_n^\prime(x) = e^{-x}\frac{d}{dx}e^xp_n(x) $$ which can then be inverted, \begin{align} p_n(x)&=\int_0^\infty e^{-t}p_{n+1}(x-t)dt\\ &=\mathbb{E}[p_{n+1}(x-X)] \end{align} where $X$ is a random variable with the exponential distribution (with rate parameter $1$), which has mean and variance equal to $1$. Then, we can iterate this procedure to get $$ p(x)=\mathbb{E}[p_n(x-X_1-X_2-\cdots-X_n)] $$ for independent exponentials $X_i$. By the central limit theorem, the random variables $$ Z_n\equiv\frac{n-X_1-X_2-\cdots-X_n}{\sqrt n} $$ converge in distribution to the standard normal distribution as $n\to\infty$. Then, setting $f_n(x)=n^{-m/2}p_n(\sqrt nx-n)$, \begin{align} \mathbb{E}[f_n(x+Z_n)]&=n^{-m/2}\mathbb{E}[p_n(\sqrt nx-X_1-X_2-\cdots-X_n)]\\ &=n^{-m/2}p(\sqrt nx)\\ &\to x^m \end{align} as $n\to\infty$. This suggests the result now. If $Z$ is a standard normal random variable, then the probabilists' Hermite polynomial satisfies the identity $\mathbb{E}[He_m(x+Z)]=x^m$. If we write $f_n(x)=He_m(x)+q_n(x)$ then, as $f_n$ and $He_m$ are both monic degree $m$ polynomials, $q_n$ is of degree less than $m$ and, \begin{align} \lim_{n\to\infty}\mathbb{E}[q_n(x+Z_n)]&=\lim_{n\to\infty}\mathbb{E}[f_n(x+Z_n)-He_m(x+Z_n)]\\ &=x^m-\mathbb{E}[He_m(x+Z)]\\ &=0. \end{align} Here, I applied the Central Limit Theorem¹ and put in the limit $\mathbb{E}[He_m(x+Z_n)]\to\mathbb{E}[He_m(x+Z)]$. If $q_n(x)=\sum_{r=0}^{m-1}c^{(n)}_rx^r$, then the coefficient of $x^r$ in $\mathbb{E}[q_n(x+Z_n)]$ is $$ c^{(n)}_r + \sum_{k=1}^{m-r-1}\binom{r+k}{k}c^{(n)}_{r+k}\mathbb{E}[Z_n^k], $$ which has to tend to $0$ as $n\to\infty$. This immediately implies that $c^{(n)}_{m-1}\to0$. As the moment $\mathbb{E}[Z_n^k]$ converge to the moments of a standard normal, they are bounded. So, once we have shown that $\lim_{n\to\infty}c^{(n)}_{r_k}=0$ for each $k > 0$, it follows that $\lim_{n\to\infty}c^{(n)}_r=0$. So, by induction on $r$, $c^{(n)}_r\to0$ as $n\to\infty$, from which the limits $q_n\to0$ and $f_n\to He_m$ follow.

¹ Above, I applied the Central Limit Theorem in the form $\mathbb{E}[g(Z_n)]\to\mathbb{E}[g(Z)]$ for polynomials $g$. This is not quite the form in which it is usually stated, where convergence in distribution is used (i.e., $g$ is a continuous bounded function). As the variables $X_n$ have finite moments, the moments of $Z_n$ do indeed converge to the moments of the normal $Z$ or, equivalently, $g$ can be taken to be any polynomial. I'll try to find a reference to the CLT in this form, although it is straightforward to prove (rather easier than the standard CLT).