$\newcommand{\ZZ}{{\mathcal Z}_G} \newcommand{\NN}{{\mathcal N}_G} \newcommand{\zz}{{\mathfrak z}_G} \newcommand{\Lie}{{\rm Lie\,}} \renewcommand{\tt}{{\mathfrak t}} \renewcommand{\gg}{{\mathfrak g}} \newcommand{\X}{{\sf X}} \newcommand{\Z}{{\Bbb Z}}$ Yes, this is true for any $n\ge 3$.
Let $G$ be a semisimple group of adjoint type over an algebraically closed field $k$ of characteristic 0. Let $T\subset G$ be a maximal torus. Write $$E=T^{(2)}:= \{t\in T(k)\ |\ t^2=1\}.$$ We wish to compute the centralizer $\ZZ(E)$ and the normalizer $\NN(E)$.
Observe that $\NN(E)\supseteq \ZZ(E)\supseteq T$. We compute $\zz(E):=\Lie \ZZ(E)$.
Lemma 1. $\zz(E)=\tt:=\Lie T$.
Proof. Since $\ZZ(E)\supseteq T$, we have $\zz(E)\supseteq\tt$. Write the root decomposition $$ \Lie G=\tt\oplus\bigoplus_{\beta\in R}\gg_\beta\,,$$ where $R=R(G,T)\subset \X^*(T)$ is the root system. Since $\zz(E)\supseteq\tt$, we have $$ \zz(E)=\tt\oplus\bigoplus_{\beta\in M}\gg_\beta$$ for some subset $M\subseteq R$. Here $$M=\{\beta\in R\ |\ \beta(t)=1\ \forall t\in E\}.$$ Let $S\subset R$ be a system of simple roots (a basis of $R$). Since $G$ is of adjoint type, the set $S\subset R\subset \X^*(T)$ is a basis of the character group $\X^*(T)$ of $T$. It follows that for any simple root $\alpha\in S$, there exists $t\in E=T^{(2)}$ such that $\alpha(t)=-1$. Let $W=W(G,T)=\NN(T)/T$ be the Weyl group. The group $W$ acts on $E$ and on $R$, and $W\cdot S=R$. Therefore, for any root $\beta\in R$, there exists $t\in E$ such that $\beta(t)=-1$, and therefore $M=\varnothing$ and $\zz(E)=\tt$, as required.
We compute $\NN(E)$. It follows from Lemma 1 that the identity component $\ZZ(E)^0$ of $\ZZ(E)$ is $T$. We see that $\NN(E)$ normalizes $\ZZ(E)$ and $\ZZ(E)^0=T$, that is, $\NN(E)\subseteq \NN(T)$. On the other hand, $\NN(T)$ normalizes $T$ and $E$, whence $\NN(T)\subseteq \NN(E)$. Thus $\NN(E)=\NN(T)$.
We wish to compute $\ZZ(E)$. Consider the Weyl group $W=\NN(T)/T$. Set $$W_E=\ZZ(E)/T\subseteq \NN(T)/T=W.$$ Then $W_E$ is a finite group, the kernel of the natural homomorphism $W\to{\rm Aut\,} E$, and $\ZZ(E)$ is the preimage of $W_E$ in $\NN(T)$.
Lemma 2. Let $W'=S_n$ (the permutation group on $n$ symbols) naturally acting on $$E'=\ker\, \Sigma\colon (\Z/2\Z)^n\to \Z/2\Z.$$ If $n\ge 3$, then this action is effective (has trivial kernel).
Proof. Left to OP and the reader.
Observe that Lemma 2 is false for $n=2$, when ${\rm Aut\,} E'=\{1\}$ whereas $S_2\neq \{1\}$.
Theorem. Let $G={\rm PGL}_n$ with $n\ge 3$. Then $\NN(E)/\ZZ(E)=\NN(T)/T=W$.
Proof. We have seen that $\NN(E)=\NN(T)$. By Lemma 2, if $n\ge 3$, then $W_E=\{1\}$, whence $\ZZ(E)=T$, and the theorem follows.