Update 1
1) I found that for $n=4, 5$, although there are infinitely many stationary points,
each stationary point is the global maximizer. In other words, the objective function is constant for all stationary points. Is this true for $n > 5$?
Take $n=4$ for example.
Let
$$g(x_1, x_2, x_3) = \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2} ,$$
\begin{align}
f(x_1, x_2, x_3) = \ln g(x_1, x_2, x_3) &= \ln \sin x_1 + \ln \sin x_2 + \ln \sin x_3 + \ln \sin (x_1 + x_2 + x_3)\nonumber\\
&\qquad - 2\ln \sin (x_1+x_2) - 2\ln \sin (x_2 + x_3).
\end{align}
The stationary points are those feasible points with $\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0.$
I found that $$\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial x_2} = \frac{\partial f}{\partial x_3} = 0 \Longrightarrow
g(x_1, x_2, x_3) = \frac{1}{4}.$$
In detail, we have
\begin{align}
\cot x_1 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) &= 0, \qquad (1)\\
\cot x_2 + \cot (x_1 + x_2 + x_3) - 2\cot (x_1 + x_2) - 2\cot (x_2 + x_3) &= 0,\\
\cot x_3 + \cot (x_1 + x_2 + x_3) - 2\cot (x_2 + x_3) &= 0.
\end{align}
By letting $u_1 =\cot x_1, \ u_2 = \cot x_2, \ u_3 = \cot x_3$,
we have $(1) \Longrightarrow u_1u_2 + u_2u_3 + u_3u_1 - u_2^2 - 2 = 0$.
On the other hand, $g(x_1, x_2, x_3) - \frac{1}{4} = \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2} = 0.$
2) We can see this from another view.
Case $n=4$: Let $u_i = \cot x_i, \ i=1,2,3$. We have (noting that $x_4 = \pi - x_1-x_2-x_3$)
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)}\nonumber\\
=\ & \frac{\sin x_1\sin x_2 \sin x_3 \sin (x_1+x_2+x_3)}{(\sin (x_1+x_2))^2 (\sin (x_2+x_3))^2}\nonumber\\
=\ & \frac{(u_1u_2 + u_2u_3+u_3u_1-1)(1+u_2^2)}{(u_1+u_2)^2(u_2+u_3)^2}\nonumber\\
=\ & \frac{1}{4} - \frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2}.
\end{align}
Case $n=5$: Let $u_i = \cot x_i, \ i=1,2,3, 4$. We have (noting that $x_5 = \pi - x_1-x_2-x_3-x_4$)
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_5)\sin (x_5+x_1)}\nonumber\\
=\ & \frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin (x_1+x_2+x_3+x_4)}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_1+x_2+x_3)\sin (x_2+x_3+x_4)}\nonumber\\
=\ & \frac{(u_1u_2u_3 + u_1u_2u_4 + u_1u_3u_4 + u_2u_3u_4 - u_1-u_2-u_3-u_4)(1+u_2^2)(1+u_3^2)}
{(u_1+u_2)(u_2+u_3)(u_3+u_4)(u_1u_2+u_2u_3+u_3u_1-1)(u_2u_3+u_3u_4+u_4u_2-1)}.\quad (2)
\end{align}
Denote (2) as $f(u_1, u_2, u_3, u_4)$.
It follows from $\frac{\partial f}{\partial u_4} = 0$
that $u_1 = g(u_2, u_3, u_4)$. Let $h(u_2, u_3, u_4) = f(g(u_2, u_3, u_4), u_2, u_3, u_4)$.
It follows from $\frac{\partial h}{\partial u_4} = 0$ that $u_2 = F(u_3, u_4)$.
Then $h(F(u_3, u_4), u_3, u_4) = \frac{5\sqrt{5}-11}{2}.$
Remark: Here $g, h, F$ are some rational functions whose expressions are not given, for the sake of simplicity.
Update
Proof of $n=4$:
We need to prove that
$$\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} \le \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4.$$
It suffices to prove that
$$(\sin(x_1+x_2))^2(\sin(x_2+x_3))^2 - 4\sin x_1 \sin x_2 \sin x_3 \sin(x_1+x_2+x_3) \ge 0.$$
Using substitutions $$\cos x_1 = \frac{1-w_1^2}{1+w_1^2}, \ \sin x_1 = \frac{2w_1}{1+w_1^2}, \ \cos x_2 = \frac{1-w_2^2}{1+w_2^2}, \ \sin x_2 = \frac{2w_2}{1+w_2^2}, \\
\cos x_3 = \frac{1-w_3^2}{1+w_3^2}, \ \sin x_3 = \frac{2w_3}{1+w_3^2},$$ the inequality becomes
$$\frac{16 Q^2}{(w_1^2+1)^2 (w_2^2+1)^4 (w_3^2+1)^2}\ge 0$$
where
\begin{align}
Q &= w_1^2 w_2^3 w_3+w_1^2 w_2^2 w_3^2-w_1 w_2^4 w_3+w_1 w_2^3 w_3^2-w_1^2 w_2^2-w_1^2 w_2 w_3-w_1 w_2^3-6 w_1 w_2^2 w_3\nonumber\\
&\qquad -w_1 w_2 w_3^2-w_2^3 w_3-w_2^2 w_3^2+w_1 w_2-w_1 w_3+w_2^2+w_2 w_3.
\end{align}
We are done.
Remark: We can prove $n=4$ without using above substitutions. For $n=5$, it is not so simple.
Previously written
This is not an answer. I want to point out that for $n=4, 5$, there exist infinitely many feasible points such that equality occurs. In other words, if the inequality holds, there exist infinitely many global maximizers.
1) $n=4$.
Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
(\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2 = 0.
\end{align}
Remark: We may solve $x_1$ from (1), that is, $x_1 = \mathrm{arccot}\frac{\sqrt{2}-\cos x_2}{\sin x_2}, \ x_2 \in (0, \frac{\pi}{2}).$
Let $x_3 = x_1,\ x_4 = \pi - x_1 - x_2 - x_3.$ We have $x_1, x_2, x_3, x_4 > 0; \ x_1 + x_2 + x_3 + x_4 = \pi$ and
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4}{\sin(x_1+x_2)\sin (x_2+x_3)\sin (x_3+x_4)\sin (x_4+x_1)} - \Big(\frac{\sin \frac{\pi}{4}}{\sin\frac{\pi}{2}}\Big)^4\\
=\ & \frac{(\sin x_1)^2\sin x_2 \sin (2x_1 + x_2)}{(\sin (x_1+x_2))^4} - \frac{1}{4}\\
=\ & - \frac{((\cot x_1)^2 + 2\cot x_1 \cot x_2 - (\cot x_2)^2 -2)^2}{4(\cot x_1 + \cot x_2)^4}\\
=\ &0.
\end{align}
2) $n = 5$.
Let $x_1, x_2 \in (0, \frac{\pi}{2})$ satisfying
\begin{align}
-4(\cot x_2)^2(\cot x_1)^2 + (-2(\cot x_2)^3 + 6\cot x_2)\cot x_1 + (\cot x_2)^4 + 4(\cot x_2)^2 - 1 = 0.
\end{align}
Let $y_1 = \cot x_1, \ y_2 = \cot x_2$. We have $-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1
+y_2^4+4y_2^2-1 = 0$ which results in $\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3 = 0$
since $x_1, x_2 \in (0, \frac{\pi}{2}).$
Let $x_3 = x_2, \ x_4 = x_1, \ x_5 = \pi - x_1 - x_2 - x_3 - x_4.$
We have $x_1, x_2, x_3, x_4, x_5 > 0; \ x_1 + x_2+x_3+x_4+x_5=\pi.$
Note that $\big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\big)^5 = \frac{5\sqrt{5}-11}{2}$. We have
\begin{align}
&\frac{\sin x_1 \sin x_2 \sin x_3 \sin x_4 \sin x_5}
{\sin (x_1+x_2) \sin (x_2 + x_3) \sin (x_3 + x_4) \sin (x_4+x_5) \sin (x_5+x_1)}
- \Big(\frac{\sin \frac{\pi}{5}}{\sin \frac{2\pi}{5}}\Big)^5\\
=\ &\frac{(\sin x_1)^2(\sin x_2)^2\sin (2x_1 + 2x_2)}{(\sin (x_1+x_2))^2\sin 2x_2 (\sin (x_1 + 2x_2))^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \frac{5\sqrt{5}-11}{2}\\
=\ &-\frac{5\sqrt{5}-11}{16}\frac{
(\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\sqrt{5}+1)(\sqrt{5}y_2^2-4y_1y_2-y_2^2+\sqrt{5}+3)^2}
{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\
=\ &0.
\end{align}
Best Answer
Proof for $p ≥ 1$
Since $u^p - 1 ≥ p(u - 1)$ for all $u ≥ 0$, it suffices to prove the result for $p = 1$. That follows from
$$\frac{x^y}{y} - 1 ≥ \frac{1 + y \ln x}{y} - 1 = \ln x + \frac1y - 1 ≥ \ln x + \ln \frac1y = \ln x - \ln y$$
by cyclic summation over $(x, y) = (x_i, x_{i + 1})$.
Conjectured proof for $p ≥ \frac12$
Since $u^p - 1 ≥ 2p(u^{\frac12} - 1)$ for all $u ≥ 0$, it suffices to prove the result for $p = \frac12$. Numerical evidence suggests that
$$\left(\frac{x^y}{y}\right)^{\frac12} - 1 ≥ \frac{\ln x}{2\sqrt[4]{1 + \frac13 \ln^2 x}} - \frac{\ln y}{2\sqrt[4]{1 + \frac13 \ln^2 y}}$$
for all $x, y > 0$. If this is true, cyclic summation yields the desired result.
Counterexample for $0 < p < \frac12$
Let $g(x) = \left(\frac{x^{1/x}}{1/x}\right)^p + \left(\frac{(1/x)^x}{x}\right)^p$. Then $g(1) = 2$, $g'(1) = 0$, and $g''(1) = 4p(2p - 1) < 0$, so we have $g(x) < 2$ for $x$ in some neighborhood of $1$. This yields counterexamples for all even $n$:
$$\left(x, \frac1x, x, \frac1x, \dotsc, x, \frac1x\right), \quad x ≈ 1, \quad 0 < p < \frac12.$$
For $n = 3$, the best counterexample seems to be
$$(0.41398215, 0.73186577, 4.77292996), \quad 0 < p < 0.39158477.$$