Now a complete proof, using some of the suggestions in the previous version of this answer. 

Note that 
\begin{equation}
G:=
F_n''(x)	x^{2-2 n} \left(x^{2 n}+1\right)^2 \left(x^{4 n}+x\right)^2 \left(x^{2 n+1}-1\right)^2
   \left(x^{2 n+1}+1\right)^2 \left(x^{2 n+2}-1\right)^2
\end{equation}
is a polynomial in $n$, $x$, and $y:=x^{2n}$, of degree $2$ in $n$:
\begin{equation}
	G=a+bn+cn^2,
\end{equation}
where $a,b,c$ are certain polynomials in $x,y$. 

The check of $G\ge0$ for $n=2,\dots,11$ is straightforward. So, assume $n\ge12$, whence $0<y<x^{24}<x^{12}$. 
We also always assume that $0<x<1$, $0<y<x^{12}$ , 
\begin{equation}
	n=\frac{\ln y}{2\ln x}; \tag{1}
\end{equation}
the latter relation is of course just another form of the equality $y=x^{2n}$. 

Note that 
\begin{equation}
	a\ge0
\end{equation}
for all $x,y$ in $(0,1)$. This is is easily verified by using Tarski's theory on quantifier elimination. In Mathematica, this theory is implemented via command Reduce[] and related commands. One can also use resultants or other tools to the same end. Details of the calculations can be seen in the  [Mathematica notebook][1] and its [pdf image][2] . 

Next, introduce 
\begin{equation}
	y_0 := 164/10^5, \quad y_1 := 695/1000,\quad x_* := 869/1000. 
\end{equation} 

The following cases are exhaustive: 

Case 1: $y\le y_1$. Then a Reduce[] quickly shows that $c\ge0$. So, for the derivative $G'_n$ of $G$ in $n$, we have 
$$G'_n=b+2cn\ge b+24c.$$
Now we have to distinguish two subcases: 

Subcase 1.1: $y\le y_1$ and ($y\ge y_0$ or $x\le x_*$). Then, using another two instances of Reduce[], with conditions $y\ge y_0$ and $x\le x_*$ respectively, we get $b + 24 c \ge 0$, so that $G'_n\ge0$, so that $G$ is increasing in $n\ge12$, and so, $G\ge G|_{n=12}$, which latter is $\ge0$, by yet another two instances of Reduce[], with conditions $y\ge y_0$ and $x\le x_*$ respectively. 
So, $G\ge0$ in Subcase 1.2. 

Subcase 1.2: $y\le y_1$ and $y<y_0$ and $x>x_*$. Then, of course, the condition $y\le y_1$ is redundant. Let here 
\begin{equation}
	\rho(x):=\frac7{2(1-x)}. 
\end{equation}
Another Reduce[] quickly shows that $b+c\rho(x)\ge0$. So, similarly to Subcase 1.1, for all $n\ge\rho(x)/2$ we get $G\ge G|_{n=\rho(x)/2}$, which latter is $\ge0$, by another Reduce[]. 
In this subcase, it remains to note that for our particular $n$, as in $(1)$, we indeed have $n\ge\rho(x)/2$. This follows because for $y<y_0$ and $x>x_*$ 
\begin{equation}
	\ln\frac1y>\ln\frac1{y_0}>1.08\frac7{2(1-x_*)}\,\ln\frac1{x_*}
	>\frac7{2(1-x)}\,\ln\frac1{x}=\rho(x)\ln\frac1{x}. 
\end{equation}
So, $G\ge0$ in Subcase 1.2 as well. 

It remains to consider 

Case 2: $y>y_1$. Then another Reduce[] yields $b\ge0$. So, if $c\ge0$, then trivially $G\ge0$ (since $a\ge0$ always). So, without loss of generality, $c<0$ and hence $G$ is concave in $n$. So, it is enough to bracket the $n$ as in $(1)$ between some simple rational expressions $n_1$ and $n_2$ such that $G|_{n=n_1}\ge0$ and $G|_{n=n_2}\ge0$. 

In Case 2, $y$ is "close" to $1$ and hence, in view of inequality $y<x^{24}$, so is $x$: $x>y_1^{1/12}>0.97$. So, it is a bit more convenient here to change variables $x, y$ to "small variables" $u:=1-x$ and $v:=1-y$. 
Then $0<v<1-y_0=0.305$ and $1-u=x>y^{1/12}=(1-v)^{1/12}>1-v/3$, whence $u<v/3$. Let now $t:=u/v$, so that $0<t<1/3$. 

The mentioned brackets $n_1$ and $n_2$ are respectively $\ell_1/2$ and $\ell_2/2$, where  
\begin{equation}
	\ell_1:=\frac1t,\quad\ell_2:=\frac{(2 - v) (1 - t v)}{t (1 - v) (2 - t v)}. 
\end{equation}
It is not hard to see (details on this are in the same Mathematica notebook) that then indeed $n_1 < n < n_2$. 
Finally, another two instances of Reduce[] show that indeed $G|_{n=n_1}\ge0$ and $G|_{n=n_2}\ge0$. 
 
The proof is complete. 


  [1]: https://www.dropbox.com/s/ne37m2co9fefpqa/solution5.nb?dl=0
  [2]: https://www.dropbox.com/s/38s2ljoj5ikds87/solution5.pdf?dl=0