Conjecture for $d=1$: Set $p:=[\frac{n}{2}]$ (integral part). Define the sequence $v_1,\ldots,v_p$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^pv_j^2=\frac12.$$ It corresponds to the projection on the line spanned by $u:=(-a_1,\ldots,-a_p, (0,) \, a_p,\ldots,a_1)$ where $a_j:=av_j$. The zero is present if $n$ is odd.

In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I$ of $n-p+1,\ldots,n$.

Asymptotically, we have $v_k\sim c k^{-1/2}$ and $a\sim c'(\log n)^{-1/2}$.

Edit. We have $v_k=\sqrt{k}-\sqrt{k-1}\sim\frac{1}{2\sqrt k}$. Therefore $a\sim\sqrt{\frac{2}{\log n}}$.

Edited. Conjecture for $d=1$: Define the sequence $v_1,\ldots,v_n$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^nv_j^2=1.$$ It corresponds to the projection on the line spanned by $u:=(a_1,\ldots,a_n)$ where $a_j:=av_j$.

In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I=(1,\ldots,p)$ with $1\le p\le n$.

We have $v_k=\sqrt{k}-\sqrt{k-1}\sim\frac{1}{2\sqrt k}$. Asymptotically, we have $a\sim\frac{2}{\sqrt{\log n}}$.

Conjecture for $d=1$: Set $p:=[\frac{n}{2}]$ (integral part). Define the sequence $v_1,\ldots,v_p$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^pv_j^2=\frac12.$$ It corresponds to the projection on the line spanned by $u:=(-a_1,\ldots,-a_p, (0,) \, a_p,\ldots,a_1)$ where $a_j:=av_j$. The zero is present if $n$ is odd.

In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I$ of $n-p+1,\ldots,n$.

Asymptotically, we have $v_k\sim c k^{-1/2}$ and $a\sim c'(\log n)^{-1/2}$.

Conjecture for $d=1$: Set $p:=[\frac{n}{2}]$ (integral part). Define the sequence $v_1,\ldots,v_p$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^pv_j^2=\frac12.$$ It corresponds to the projection on the line spanned by $u:=(-a_1,\ldots,-a_p, (0,) \, a_p,\ldots,a_1)$ where $a_j:=av_j$. The zero is present if $n$ is odd.

In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I$ of $n-p+1,\ldots,n$.

Asymptotically, we have $v_k\sim c k^{-1/2}$ and $a\sim c'(\log n)^{-1/2}$.

Edit. We have $v_k=\sqrt{k}-\sqrt{k-1}\sim\frac{1}{2\sqrt k}$. Therefore $a\sim\sqrt{\frac{2}{\log n}}$.