Note: this a trial approach via the Gamma function.
Looks promising but at the moment I cannot find a way to close the last step.
Maybe someone can help to conclude ?
Since we are given
$$
\left\{ \matrix{
0 < x_{\,k} < \pi \hfill \cr
x_{\,1} + x_{\,2} + \cdots + x_{\,n} = \pi \hfill \cr
{{\sin x_{\,1} \sin x_{\,2} \cdots \sin x_{\,n} } \over {\sin \left( {x_{\,1} + x_{\,2} } \right)\sin \left( {x_{\,1} + x_{\,2} } \right) \cdots \sin \left( {x_{\,n} + x_{\,1} } \right)}}
\le {{\sin \left( {\pi /n} \right)^{\,n} } \over {\sin \left( {2\pi /n} \right)^{\,n} }} \hfill \cr} \right.
$$
and since it is
$$
\Gamma \left( z \right)\,\Gamma \left( {1 - z} \right) = {\pi \over {\sin \left( {\pi \,z} \right)}}\quad \left| {\;\forall z \in \mathbb C\backslash \mathbb Z} \right.
$$
Then, putting
$$
x_{\,k} = \pi \,z_{\,k} \quad \left| {\;0 < z_{\,k} < 1} \right.
$$
so $z_k \notin \mathbb Z$ and we can apply the relation with Gamma to obtain
$$
\eqalign{
& {{\sin x_{\,1} \sin x_{\,2} \cdots \sin x_{\,n} } \over {\sin \left( {x_{\,1} + x_{\,2} } \right)\sin \left( {x_{\,1} + x_{\,2} } \right)
\cdots \sin \left( {x_{\,n} + x_{\,1} } \right)}} = \cr
& = {{\Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)}
\over {\Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)}}\; \cdot \cr
& \cdot \;{{\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right)}
\over {\Gamma \left( {1 - z_{\,1} } \right)\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right)}} = \cr
& = {{z_{\,1} ^{\,\overline {\,z_{\,2} \,} } z_{\,2} ^{\,\overline {\,z_{\,3} \,} } \cdots z_{\,n} ^{\,\overline {\,z_{\,1} \,} } }
\over {\left( {1 - z_{\,1} - z_{\,2} } \right)^{\,\overline {\,z_{\,2} \,} } \left( {1 - z_{\,2} - z_{\,3} } \right)^{\,\overline {\,z_{\,3} \,} }
\cdots \left( {1 - z_{\,n} - z_{\,1} } \right)^{\,\overline {\,z_{\,1} \,} } }} = \cr
& = {{z_{\,1} ^{\,\overline {\,z_{\,2} \,} } z_{\,2} ^{\,\overline {\,z_{\,3} \,} } \cdots z_{\,n} ^{\,\overline {\,z_{\,1} \,} } }
\over {\left( { - z_{\,1} } \right)^{\,\underline {\,\,z_{\,2} \,} } \left( { - z_{\,2} } \right)^{\,\underline {\,\,z_{\,2} \,} }
\cdots \left( { - z_{\,n} } \right)^{\,\underline {\,\,z_{\,2} \,} } }} \cr}
$$
where $z^{\,\underline {\,w\,} } ,\quad z^{\,\overline {\,w\,} } $ represent respectively the Falling and Rising Factorial
It is in fact known that
$$
\left( { - z} \right)^{\;\underline {\,w\,} } = {{\,\sin \left( {\pi \,\left( {z + w} \right)} \right)} \over {\sin \left( {\pi \,z} \right)\,}}\;z^{\,\overline {\,w\,} }
$$
which corresponds to the "upper negation" identity for complex Binomials.
However, for our scope, we have better and restart from the second line and use the fact that $ln \Gamma(z)$ is convex,
and decreasing in $(0,1)$.
That means that the average of the log of gamma at a number of points is no less than the log of gamma taken at the average point, or
$$
\Gamma \left( {{1 \over n}} \right)^{\,n} \le \Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)
$$
and consequently
$$
\eqalign{
& \Gamma \left( {1 - {1 \over n}} \right)^{\,n} \le \Gamma \left( {1 - z_{\,1} } \right)\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right) \cr
& \Gamma \left( {{2 \over n}} \right)^{\,n} \le
\Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right) \cr
& \Gamma \left( {1 - {2 \over n}} \right)^{\,n} \le
\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \cr}
$$
Therefore
$$
\eqalign{
& \Gamma \left( {{2 \over n}} \right)^{\,n} \Gamma \left( {1 - {2 \over n}} \right)^{\,n} = \left( {{\pi \over {\sin \left( {\pi \,2/n} \right)}}} \right)^{\,n} \le \cr
& \le \Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)\; \cdot \cr
& \cdot \;\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots \Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \cr
& \Gamma \left( {{1 \over n}} \right)^{\,n} \Gamma \left( {1 - {1 \over n}} \right)^{\,n} = \left( {{\pi \over {\sin \left( {\pi \,/n} \right)}}} \right)^{\,n} \le \cr
& \le \Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)\; \cdot \Gamma \left( {1 - z_{\,1} } \right)
\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right) \cr}
$$
It remains to demonstrate that by dividing the two we get the correct inequality,
i.e. that it is
$$
\eqalign{
& 1 \le {{\left( \matrix{
\Gamma \left( {z_{\,1} + z_{\,2} } \right)\Gamma \left( {z_{\,2} + z_{\,3} } \right) \cdots \Gamma \left( {z_{\,n} + z_{\,1} } \right)\; \cdot \hfill \cr
\cdot \;\Gamma \left( {1 - z_{\,1} - z_{\,2} } \right)\Gamma \left( {1 - z_{\,2} - z_{\,3} } \right) \cdots
\Gamma \left( {1 - z_{\,n} - z_{\,1} } \right) \hfill \cr} \right)} \over {\left( {{\pi \over {\sin \left( {\pi \,2/n} \right)}}} \right)^{\,n} }} \le \cr
& \le {{\Gamma \left( {z_{\,1} } \right)\Gamma \left( {z_{\,2} } \right) \cdots \Gamma \left( {z_{\,n} } \right)\; \cdot \Gamma \left( {1 - z_{\,1} } \right)
\Gamma \left( {1 - z_{\,2} } \right) \cdots \Gamma \left( {1 - z_{\,n} } \right)} \over {\left( {{\pi \over {\sin \left( {\pi \,/n} \right)}}} \right)^{\,n} }} \cr}
$$
That reads that the average of $\ln \Gamma(z_k + z_{k+1})$ is less dispersed around $\ln \Gamma(z_{2\, avg})$ than
$\ln \Gamma(z_k)$ around $\ln \Gamma(z_{1\, avg})$,
which intuitively makes sense (for large $n$ at least).
It remains to express that in a formal way.