$\newcommand{\al}{\alpha}$You are citing Szego's book incorrectly. The correct citation is this: for every real $\al$ and every real $a>0$, \begin{equation*} \max_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| \sim n^{\al/2-1/12} \tag{1}\label{1} \end{equation*} as $n\to\infty$.

There can be no corresponding result with $\min$ in place of $\max$ -- because, for large enough $n$, the polynomial $L_n^{(\al)}$ oscillates, so that for every real $\al$, every real $a>0$, and all large enough $n$
\begin{equation*} \min_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| =0. \tag{2}\label{2} \end{equation*}

Indeed, by Theorem 6.31.3 (p. 129) of G. Szego, Orthogonal polynomials, 1939, for $\al>-1$ and the $\nu$th smallest root $x_{\nu n}$ of $L_n^{(\al)}(x)$ we have \begin{equation*} x_{\nu n}>\frac{(j_\nu/2)^2}{n+(\al+1)/2} \end{equation*} where $\nu=1,\dots,n$ and, according to the first sentence after that theorem, $(j_\nu/2)^2\sim\pi^2\nu^2/4$ as $\nu\to\infty$. So, for any fixed natural $k$, uniformly over all $\nu\in\{n-k+1,\dots,n\}$ we have $x_{\nu n}\gtrsim\frac{\pi^2}4\,n$.

Therefore and because $\frac{\pi^2}4>1$, for any real $\al>-1$, any fixed natural $k$, and all large enough $n$, there are at least $k$ roots of $L_n^{(\al)}(x)$ in the interval $[n,\infty)$, and hence \eqref{2} holds.

By Rolle's theorem and the formula \begin{equation*} \frac d{dx}\,[x^\al e^{-x}\,L_n^{(\al)}(x)]=(n+1)x^{\al-1} e^{-x}\,L_{n+1}^{(\al-1)}(x) \end{equation*} at the top of page 287, it follows that for any fixed natural $k$, any real $\al>-1$, and all large enough $n$, there are at least $k-1$ roots of $L_{n+1}^{(\al-1)}(x)$ in the interval $[n,\infty)$. That is, for any fixed natural $k$, any real $\al>-1-1$, and all large enough $n$, there are at least $k-1$ roots of $L_n^{(\al)}(x)$ in the interval $[n-1,\infty)$. Continuing thus, we conclude that, for any fixed natural $k$, any real $\al>-1-(k-1)$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$. That is, for any real $\al$, any fixed natural $k>-\al$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$.

Thus, we have \eqref{2} for every real $\al$, every real $a>0$, and all large enough $n$.

$\newcommand{\al}{\alpha}$You are citing Szegő's book incorrectly. The correct citation is this: for every real $\al$ and every real $a>0$, \begin{equation*} \max_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| \sim n^{\al/2-1/12} \tag{1}\label{1} \end{equation*} as $n\to\infty$.

There can be no corresponding result with $\min$ in place of $\max$ -- because, for large enough $n$, the polynomial $L_n^{(\al)}$ oscillates, so that for every real $\al$, every real $a>0$, and all large enough $n$
\begin{equation*} \min_{x\ge a}e^{-x/2} x^{\al/2+1/4}|L_n^{(\al)}(x)| =0. \tag{2}\label{2} \end{equation*}

Indeed, by Theorem 6.31.3 (p. 129) of G. Szegő, Orthogonal polynomials, 1939, for $\al>-1$ and the $\nu$th smallest root $x_{\nu n}$ of $L_n^{(\al)}(x)$ we have \begin{equation*} x_{\nu n}>\frac{(j_\nu/2)^2}{n+(\al+1)/2} \end{equation*} where $\nu=1,\dots,n$ and, according to the first sentence after that theorem, $(j_\nu/2)^2\sim\pi^2\nu^2/4$ as $\nu\to\infty$. So, for any fixed natural $k$, uniformly over all $\nu\in\{n-k+1,\dots,n\}$ we have $x_{\nu n}\gtrsim\frac{\pi^2}4\,n$.

Therefore and because $\frac{\pi^2}4>1$, for any real $\al>-1$, any fixed natural $k$, and all large enough $n$, there are at least $k$ roots of $L_n^{(\al)}(x)$ in the interval $[n,\infty)$, and hence \eqref{2} holds.

By Rolle's theorem and the formula \begin{equation*} \frac d{dx}\,[x^\al e^{-x}\,L_n^{(\al)}(x)]=(n+1)x^{\al-1} e^{-x}\,L_{n+1}^{(\al-1)}(x) \end{equation*} at the top of page 287, it follows that for any fixed natural $k$, any real $\al>-1$, and all large enough $n$, there are at least $k-1$ roots of $L_{n+1}^{(\al-1)}(x)$ in the interval $[n,\infty)$. That is, for any fixed natural $k$, any real $\al>-1-1$, and all large enough $n$, there are at least $k-1$ roots of $L_n^{(\al)}(x)$ in the interval $[n-1,\infty)$. Continuing thus, we conclude that, for any fixed natural $k$, any real $\al>-1-(k-1)$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$. That is, for any real $\al$, any fixed natural $k>-\al$, and all large enough $n$, there is at least one root of $L_n^{(\al)}(x)$ in the interval $[n-k,\infty)$.

Thus, we have \eqref{2} for every real $\al$, every real $a>0$, and all large enough $n$.