Not a complete proof, but some suggestions. 

Note that 
\begin{equation}
G(x):=
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 $n$. We have $0<x<1$, $0<y\le x^4$, $n=\frac{\ln y}{2\ln x}$. Letting $t:=y/x^4$, we have $0<t<1$. Accordingly replacing $n$ by $\frac{\ln y}{2\ln x}$ and then $y$ by $tx^4$, rewrite $G(x)\ln^2 x$ as the following polynomial P in $x,t,\ln x,\ln t$: 

[![enter image description here][1]][1]

This polynomial P is quadratic in $\ln x,\ln t$. We need to show that P$\ge0$ for $x$ and $t$ in $(0,1)$. 
(Numerical minimization suggests it is so.) By Tarski's theory, this can be done purely algorithmically, because for any polynomial $p(x)$ of degree $d$ and any natural $n$ one has 
$x^{d+1}\big(p(x)\ln^n x\big)^{(d+1)}$ is a polynomial in $x$ and $\ln x$ of degree $n-1$ in $\ln x$, so that the exponent of $\ln x$ gets decreased by $1$. This way, repeated differentiation and multiplication by powers of $x$ will kill off all the logarithms, leaving us only with polynomials. However, the computational complexity of this approach in this case seems overwhelming. 

This procedure can be sped up sometimes. In particular, in our case the coefficient (say $a$) of $\ln^2x$ in P seems positive everywhere. So, dividing P by $a$ and then differentiating $P/a$ in $x$, one gets rid of $\ln^2x$ in just one step. 
Still, the complexity seems overwhelming. 

Yet another way to kill the logarithms is as follows. It is enough to show that P does not vanish on the square $(0,1)^2$. Since P is quadratic in $\ln x$ and $\ln t$, we can solve equation P$=0$ for $r:=\frac{\ln t}{\ln x}$, obtaining two roots, say $r_\pm$, of the quadratic equation. (A bonus (?) that we get here is the extra condition that the discriminant of the quadratic equation be nonnegative.) 
Note that $r_\pm$ are algebraic functions of $x,t$, in radicals of order $2$. It is then enough to show that $d:=\ln t-r_\pm\ln x\ne0$ on $(0,1)^2$. Here, $d$ is linear in $\ln t$ and $\ln x$, and so, this logarithms can be killed much faster. However, even though $r_\pm$ are algebraic functions of $x$ and $t$, in radicals of order $2$ -- they are not polynomial. So, the complexity again seems overwhelming.   

Perhaps the best approach here is the interval method. Namely, partition the square $(0,1)^2$, where the point $(x,t)$ lies, into a grid of smaller squares. Within each small square, bound each "term" in P tightly enough from below to show that P$\ge0$ on each such small square. (We want those "terms" to be as big and few as possible.) Then we will be done. 

In principle, the only problematic small squares in this method are the ones of one of the following two kinds: (i) adjacent to the lines $x=0$ or $t=0$, where $\ln x$ or $\ln t$ are singular and (ii) where P vanishes at some point of the boundary of the small square. For the log-"singular" small squares of type (i), we can use the described above killing of $\ln x$ and/or $\ln t$, which is easier to control in small squares. For the small squares of type (ii), we can e.g. use a controlled Taylor expansion. For the small squares that are both of type (i) and type (ii), we can use a combination of the respective methods.


  [1]: https://i.sstatic.net/ws2Fy.png