$\newcommand{\thh}{\theta} \newcommand{\ep}{\varepsilon} $ Let us show that the spectral gap is on the order of at least $1/n^3$.

In what follows, $z$ always denotes a root of the equation \begin{equation*} f(z):=z^n-pz^{n-1}-q=0, \end{equation*} so that $|z|\le1$.

Let $c$, possibly with indices, denote various positive expressions (possibly different even within one formula) which stay away from both $0$ and $\infty$ as $n\to\infty$.

Lemma 1. $|z|\ge1-c/n$.

Proof. We have $q\le|z|^n+p|z|^{n-1}\le2|z|^{n-1}$, whence $|z|\ge(\frac q2)^{1/(n-1)}=1-c/n$. $\Box$

Lemma 2. Suppose that $z=|z|e^{i\thh}$ and $0<|\thh|\le\pi$. Then $|\thh|\ge1/n$ eventually (for large enough $n$).

Proof. Let $\ep:=(1-p)/2>0$. If $\cos\thh\le p+\ep[<1]$, then $|\thh|\ge c\ge1/n$ eventually, as desired. So, without loss of generality (wlog) $\cos\thh>p+\ep$. So, eventually $\cos\thh-p/|z|>p+\ep-p/|z|>\ep/2$ by Lemma 1, and hence \begin{equation*} e^{i\thh}-p/|z|=\cos\thh-p/|z|+i\sin\thh=re^{i\phi} \tag{0} \end{equation*} for some $r>0$ and $\phi\in(-\pi/2,\pi/2)$ such that \begin{equation*} \tan\phi=\frac{\sin\thh}{\cos\thh-p/|z|}=c\sin\thh\quad\text{and hence}\quad\phi=c\thh. \end{equation*} Note that \begin{equation*} q=z^n-pz^{n-1}=|z|^n e^{i(n-1)\thh}(e^{i\thh}-p/|z|) =|z|^n re^{i[(n-1)\thh+\phi]} \end{equation*} by (0), whence for some integer $k$ we have \begin{equation*} 2\pi k=(n-1)\thh+\phi=(n-1+c)\thh. \tag{1} \end{equation*} If $k=0$, then it would follow from (1) that $\thh=0$, which would contradict the condition $0<|\thh|\le\pi$ of Lemma 2. So, $|k|\ge1$, and Lemma 2 follows from (1). $\Box$

Lemma 3. $z\notin[0,1)$.

Proof. We have $f'(x)=nx^{n-2}(x-x_*)$, where $x_*:=\frac{n-1}n\,p$. So, $f$ is decreasing on $[0,x_*]$ and increasing on $[x_*,1]$. Also, $f(0)=-q<0$ and $f(1)=0$. Now Lemma 3 follows. $\Box$

So, Lemma 2 allows us to relax the condition $0<|\thh|$ in Lemma 3 to $z\ne1$ and thus immediately get

Lemma 2a. Suppose that $z=|z|e^{i\thh}\ne1$ and $|\thh|\le\pi$. Then $|\thh|\ge1/n$ eventually.

Now we are ready to prove the final result:

Theorem. If $z\ne1$, then $|z|\le1-c/n^3$.

Proof. Wlog $|z|>1-c/n^3$. Let $x:=\Re z$. By Lemma 2a, $x\le\cos\thh=1-c/n^2$ and hence \begin{equation} |z-p|^2=|z|^2+p^2-2px\ge(1-c/n^3)^2+p^2-2p(1-c/n^2)=q^2+c/n^2, \end{equation} and so, $|z-p|\ge q+c/n^2$. Thus, \begin{equation} |z|^{n-1}=\frac q{|z-p|}\le\frac q{q+c/n^2}=\frac1{1+c/n^2}=1-c/n^2, \end{equation} which yields the theorem. $\Box$

$\newcommand{\thh}{\theta} \newcommand{\ep}{\varepsilon} $ Let us show that the spectral gap is on the order of at least $1/n^3$.

In what follows, $z$ always denotes a root of the equation \begin{equation*} f(z):=z^n-pz^{n-1}-q=0, \end{equation*} so that $|z|\le1$, since the matrix is stochastic.

Let $c$, possibly with indices, denote various positive expressions (possibly different even within one formula) which stay away from both $0$ and $\infty$ as $n\to\infty$.

Lemma 1. $|z|\ge1-c/n$.

Proof. We have $q\le|z|^n+p|z|^{n-1}\le2|z|^{n-1}$, whence $|z|\ge(\frac q2)^{1/(n-1)}=1-c/n$. $\Box$

Lemma 2. Suppose that $z=|z|e^{i\thh}$ and $0<|\thh|\le\pi$. Then $|\thh|\ge1/n$ eventually (for large enough $n$).

Proof. Let $\ep:=(1-p)/2>0$. If $\cos\thh\le p+\ep[<1]$, then $|\thh|\ge c\ge1/n$ eventually, as desired. So, without loss of generality (wlog) $\cos\thh>p+\ep$. So, eventually $\cos\thh-p/|z|>p+\ep-p/|z|>\ep/2$ by Lemma 1, and hence \begin{equation*} e^{i\thh}-p/|z|=\cos\thh-p/|z|+i\sin\thh=re^{i\phi} \tag{0} \end{equation*} for some $r>0$ and $\phi\in(-\pi/2,\pi/2)$ such that \begin{equation*} \tan\phi=\frac{\sin\thh}{\cos\thh-p/|z|}=c\sin\thh\quad\text{and hence}\quad\phi=c\thh. \end{equation*} Note that \begin{equation*} q=z^n-pz^{n-1}=|z|^n e^{i(n-1)\thh}(e^{i\thh}-p/|z|) =|z|^n re^{i[(n-1)\thh+\phi]} \end{equation*} by (0), whence for some integer $k$ we have \begin{equation*} 2\pi k=(n-1)\thh+\phi=(n-1+c)\thh. \tag{1} \end{equation*} If $k=0$, then it would follow from (1) that $\thh=0$, which would contradict the condition $0<|\thh|\le\pi$ of Lemma 2. So, $|k|\ge1$, and Lemma 2 follows from (1). $\Box$

Lemma 3. $z\notin[0,1)$.

Proof. We have $f'(x)=nx^{n-2}(x-x_*)$, where $x_*:=\frac{n-1}n\,p$. So, $f$ is decreasing on $[0,x_*]$ and increasing on $[x_*,1]$. Also, $f(0)=-q<0$ and $f(1)=0$. Now Lemma 3 follows. $\Box$

So, Lemma 2 allows us to relax the condition $0<|\thh|$ in Lemma 3 to $z\ne1$ and thus immediately get

Lemma 2a. Suppose that $z=|z|e^{i\thh}\ne1$ and $|\thh|\le\pi$. Then $|\thh|\ge1/n$ eventually.

Now we are ready to prove the final result:

Theorem. If $z\ne1$, then $|z|\le1-c/n^3$.

Proof. Wlog $|z|>1-c/n^3$. Let $x:=\Re z$. By Lemma 2a, $x\le\cos\thh=1-c/n^2$ and hence \begin{equation} |z-p|^2=|z|^2+p^2-2px\ge(1-c/n^3)^2+p^2-2p(1-c/n^2)=q^2+c/n^2, \end{equation} and so, $|z-p|\ge q+c/n^2$. Thus, \begin{equation} |z|^{n-1}=\frac q{|z-p|}\le\frac q{q+c/n^2}=\frac1{1+c/n^2}=1-c/n^2, \end{equation} which yields the theorem. $\Box$