$\newcommand{\1}{\mathbf 1}\newcommand{\ep}{\varepsilon}\newcommand{\tr}{\operatorname{tr}}$The min-max value is $\sqrt n$.
Indeed, take any real $n\times n$ matrix $H$ with $|\det H|=1$. By the singular value decomposition,
\begin{equation}
H=U^TDV,
\end{equation}
where $U$ and $V$ are some orthogonal matrices and $D$ is the diagonal matrix with diagonal entries $d_1,\dots,d_n$ such that $d_1\cdots d_n=1$. Then
\begin{equation}
\max_{\|x\|_2\le1}\|Hx\|_1=\max_{\|z\|_2\le1}\|U^TDz\|_1.
\end{equation}
So,
\begin{equation}
\begin{aligned}
&\min_{H\colon\,|\det H|=1}\max_{\|x\|_2\le1}\|Hx\|_1 \\
&=\min_{U,D}\max_{\|z\|_2\le1}\,\max_{\ep\in\{-1,1\}^n}\sum_{i=1}^n e_i^T D_\ep U^TDz \\
&=\min_{U,D}\max_{\ep\in\{-1,1\}^n}\,\max_{\|z\|_2\le1}\,\sum_{i=1}^n e_i^T D_\ep U^TDz \\
&=\min_{U,D}\max_{\ep\in\{-1,1\}^n}\,\|DUD_\ep\1\|_2,
\end{aligned}
\end{equation}
where (i) $\min_{U,D}$ denotes the minimum over all orthogonal matrices $U$ and all diagonal matrices $D$ with diagonal entries $d_1,\dots,d_n$ such that $d_1\cdots d_n=1$; (ii) the $e_i$'s are the standard basis vectors; (iii) $\ep:=(\ep_1,\dots,\ep_n)\in\{-1,1\}^n$ and $D_\ep$ is the diagonal matrix with diagonal entries $\ep_1,\dots,\ep_n$; and (iv) $\1:=\sum_{i=1}^n e_i$.
Now the crucial point: considering $\ep$ as a random point uniformly distributed on $\{-1,1\}^n$, we get the expected value of $E\|DUD_\ep\1\|_2^2$:
\begin{equation}
E\|DUD_\ep\1\|_2^2=\1^T ED_\ep U^T D^2 U D_\ep \1
=\tr U^T D^2 U=\tr D^2=\sum_{i=1}^n d_i^2\ge n,
\end{equation}
since $d_1\cdots d_n=1$. Hence,
\begin{equation}
\max_{\ep\in\{-1,1\}^n}\,\|DUD_\ep\1\|\ge
E\|DUD_\ep\1\|_2\ge\sqrt n.
\end{equation}
So,
\begin{equation}
\min_{H\colon\,|\det H|=1}\max_{\|x\|_2\le1}\|Hx\|_1\ge\sqrt n.
\end{equation}
On the other hand,
\begin{equation}
\min_{H\colon\,|\det H|=1}\max_{\|x\|_2\le1}\|Hx\|_1\le
\max_{\|x\|_2\le1}\|x\|_1=\sqrt n.
\end{equation}
Thus,
\begin{equation}
\min_{H\colon\,|\det H|=1}\max_{\|x\|_2\le1}\|Hx\|_1=\sqrt n,
\end{equation}
as claimed.
