$\newcommand{\s}{\overset{\text{sgn}}=}
\newcommand{\Dx}{\text{Dx}}
\newcommand{\logDx}{\text{logDx}}
\newcommand{\DlogDx}{\text{DlogDx}}
\newcommand{\DDDlogDx}{\text{DDDlogDx}}
\newcommand{\DDDDDlogDx}{\text{DDDDDlogDx}}
\newcommand{\dif}{\text{dif}}
\newcommand{\Ddif}{\text{Ddif}}
\newcommand{\R}{\mathbb{R}}$
Let us show that the inequality in question holds for all real $n\ge5$; the cases when $n\in\{1,2,3,4\}$ are verified directly. By a comment of Pietro Mayer, without loss of generality $0<x<1$. We shall reduce the problem to the completely algorithmic problem of checking sign patterns of several polynomials in $n,x$, of total degrees $\le11$. This reduction is done in a few steps:

**Step 1: Eliminating $(\frac{1+x}2)^n$:**
The inequality in question can be rewritten as
\begin{equation}
u(x):=u_n(x):=n \ln \left(\frac{x^n+1}{x^{n-1}+1}\right)
-\ln \left(x^n+1-z^n\right)\ge0,
\end{equation}
where $z:=z_x:=\frac{1+x}2$.
Note that
\begin{multline*}
u'(x)\frac{x (1+x)}n \left(x^{1-n}+x^n+x+1\right) \left(x^n+1-z^n\right) \\
=\Dx:=\left(n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)\right) z^n-(n-1)
\left(1-x^2\right) \left(1+x^n\right),
\end{multline*}
so that
\begin{equation}
u'(x)\s\Dx\s\logDx(x),
\end{equation}
where $\s$ denotes the equality in sign and

\begin{equation}
\logDx(x):=\logDx_n(x):=n \ln z-\ln \frac{(n-1) \left(1-x^2\right)
\left(1+x^n\right)}{n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)}.
\end{equation}
Here and in the sequel, $\Dx$, $\logDx$, etc. are atomic, "indivisible" symbols; $\Dx$ refers to the derivative (of $u$) in $x$, $\logDx$ refers to a certain kind of logarithmic modification of $\Dx$, etc.
Next, let
\begin{multline*}
\DlogDx(x):=\DlogDx_n(x):= \\
\logDx'(x)(1-x) (1+x) x^{n-1} \left(1+x^n\right) \left(n \left(1-x^2\right)+\left(x^{2-n}-1\right)
\left(1+x^n\right)\right) \\
=n^2 (x-1)^2 (x+1) \left(x-x^n\right) x^{n-2}-2 \left(x^n-1\right) \left(x^n+1\right)^2+\frac{n
(x-1) \left(x^n+1\right)^2 \left(x^n+x\right)}{x}.
\end{multline*}
So, we get a polynomial in $x^n$ of degree $3$ over the field $\R(n,x)$ of all real rational functions in $n,x$.

**Step 2: Reducing the degree from $3$ to $2$:**
Let
\begin{multline*}
\DDDlogDx(x):= \DlogDx''(x) x^{3 - 3 n}\\
=x^{3-3 n} (n (n x-n+2 x+2) (n^2 x^2-n^2+n x^2+2 n-1) x^{n-3} \\
-2 (n-1) n (x-1)
(2 n^2 x^2-2 n^2+n x^2-2 n x+3 n+2 x) x^{2 n-4}+n (3 n-1) (3 n x-3 n-6 x+2) x^{3
n-3}) \\
\s\DlogDx''(x).
\end{multline*}
Taking the second derivative $\DlogDx''(x)$ of the polynomial $\DlogDx(x)$ in $x^n$ of over $\R(n,x)$ kills the free term of that polynomial. Thus, we get the polynomial $\DDDlogDx(x)$ of degree $2$ in $x^{-n}$ of over $\R(n,x)$.

**Step 3: Reducing the degree from $2$ to $1$:** Let
\begin{equation}
\DDDDDlogDx(x):= \frac{\DDDlogDx''(x)}{2 (n - 1) n^2 x^{-3 - 2 n}}
= A_n(x) - x^n B_n(x),
\end{equation}
\begin{equation}
A_n(x):=\left(2 n^3+3 n^2-5 n-6\right) x^4+\left(-2 n^3+3 n^2+3 n-2\right) x^3+\left(-2 n^3-n^2+5
n-2\right) x^2+\left(2 n^3-5 n^2+n+2\right) x,
\end{equation}
\begin{equation}
B_n(x):=2 n^3+3 n^2-\left(-2 n^3+5 n^2-n-2\right) x^3-\left(2 n^3+n^2-5 n+2\right) x^2-\left(2 n^3-3
n^2-3 n+2\right) x-5 n-6,
\end{equation}
so that
\begin{equation}
\DDDlogDx''(x)\s A_n(x) - x^n B_n(x).
\end{equation}
Thus, we get the polynomial $\DDDDDlogDx(x)$ of degree $1$ in $x^n$ of over $\R(n,x)$.

**Step 4: Reducing the degree from $1$ to $0$:**

We can see that (under the conditions $n\ge5$ and $0<x<1$, assumed everywhere here) $B_n(x)>0$. So, $\DDDlogDx''(x)<0$ whenever $A_n(x)\le0$.

Further, let
\begin{equation}
\dif(x) = \dif_n(x) :=\ln\frac{A_n(x)}{B_n(x)} - n \ln x\s A_n(x) - x^n B_n(x)\s \DDDlogDx''(x)
\end{equation}
wherever $A_n(x)>0$, and then
\begin{multline*}
\Ddif(x) = \Ddif_n(x) :=\dif'(x)\frac{A_n(x)B_n(x)}{(n+1)(n-2)} \\
=-4 n^5 (x-1)^4 (x+1)^2+4 n^4 (x-1)^4 (x+1)^2+n^3 (x-1)^2 \left(15 x^4+16 x^3-10 x^2+16
x+15\right) \\
-4 n^2 \left(x^2-1\right)^2 \left(5 x^2+x+5\right)+n \left(-x^6+30 x^5+41 x^4-44
x^3+41 x^2+30 x-1\right) \\
+2 \left(3 x^6-6 x^5-11 x^4-36 x^3-11 x^2-6 x+3\right) \\
\s\dif'(x),
\end{multline*}
finally getting a polynomial in $n,x$.

Now we need to trace the above steps back:

Looking back at the polynomial $A_n(x)$, (for $x\in(0,1)$) we find that $A_n(x)\le0$ iff $x_1\le x\le x_2$, where $x_1=x_1(n)$ and $x_2=x_2(n)$ are the two roots of $A_n(x)$ in $(0,1)$ such that $x_1<x_2$.

Further, $\Ddif<0$ and hence $\dif'<0$ on $(0,x_1]$;
and $\Ddif>0$ and hence $\dif'>0$ on $[x_2,1)$.
So, $\dif$ decreases on $(0, x_1]$ and increases on $[x_2, 1)$.
So, $\dif$ is $+-$ on $(0, x_1]$ (that is, $\dif$ can switch sign at most once on $(0, x_1]$, and only from $+$ to $-$). Similarly, $\dif$ is $-+$ on $[x_2, 1)$.

But also $\dif(1)=0$. So, actually $\dif<0$ on $[x_2, 1)$.

So, $\DDDlogDx''$ is $+-$ on $(0, x_1]$ and $\DDDlogDx'' < 0$ on $[x_2, 1)$.

Also, $A < 0$ and hence $\DDDlogDx'' < 0$ on $[x_1, x_2]$.
So, $\DDDlogDx''$ is $+-$ on $(0, 1)$.
So, $\DDDlogDx$ is convex-concave on $(0, 1)$.
Also, $\DDDlogDx(1)=0$.

So, $\DDDlogDx$ is $+-+$ on $(0, 1)$. So, $\DlogDx$ is convex-concave-convex on $(0, 1)$.
Also, $\DlogDx(1)=\DlogDx'(1)=\DlogDx''(1)=0>-8n(n^2-1)=\DlogDx'''(1)$ and $\DlogDx(0+)=2-n<0$.
So, $\DlogDx$ is $-+$; so, $\logDx$ is decreasing-increasing.

Also, $\logDx(1-)=0$. So, $\logDx$ is $+-$, and hence so is $\Dx$ (with $z = \frac{1 + x}2$).

Recalling that $u'(x)\s\Dx$, we see that $u_n(x)$ is increasing-decreasing (in $x\in(0,1)$). Also, $u_n(0)=-\ln(1 - 2^{-n})>0$ and $u_n(1)=0$.

Thus, $u>0$, which concludes the proof.