$\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. $\Box$