The following should give a solution. Fix $n$, and let $V(x)$ be the polynomial from the question, for which we want to show that it is positive on the open interval $(0,1)$. Set \begin{align*} a &= 64n^2 - 256/3n + 20\\ b &= 768n^3 - 1024n^2 + 272n - 24\\ c &= 14336/3n^4 - 306304/45n^3 + 35792/15n^2 - 6104/15n + 3 \end{align*} Then \begin{equation} \frac{V(x)}{(1-x^2)^4x^{2n-2}}=(1-x)^3H(x)+(n+1)^2(2n+1)^2x^{22n-4} (ax^2+bx(1-x)+c(1-x)^2) \end{equation} for a polynomial $H(x)$.
Here $a,b,c$ are positive for $n\ge2$. The point is that all coefficients of $H(x)$ are positive! I have numerically verified that for all $n\le10000$.
I'm not sure about the easiest kind to prove the assertion about the coefficients of $H(x)$. But the following should work: Multiplying the equation by $(1-x)^4(1+x)^7$ yields \begin{equation} (1-x^2)^7H(x)=W(x) \end{equation} where $W(x)$ is a polynomial for which we have an explicit (though messy) expression. In fact for $n\ge8$ $W(x)$ has exactly $128$ terms. Now the coefficients of $H(x)$ can be computed by multiplying $W(x)$ and the power series of $\frac{1}{(1-x^2)^7}$. More precisely, the following holds (verified by symbolic computation):
Consider the integer pairs $(i,j)$ where $0\le i\le 11$ and $-4\le j\le 11$. To each such pair there is a polynomial $a_{i,j}(n)$ in the variable $n$ of degree $0,1,2,8$, or $a_{i,j}(n)=0$ (for instance if $i=0$ and $j<0$). Then \begin{equation} W = \sum_{i,j}a_{i,j}(n)x^{2ni+j}. \end{equation} Thus we can write $W(x)=W_0(x^2)+xW_1(x^2)$, according to whether $j$ is even or odd. It remains to show that the coefficients of $\frac{W_0(x)}{(1-x)^7}$ and $\frac{W_1(x)}{(1-x)^7}$ are positive.
if $n\ge8$, then $2ni+j>2ni'+j'$ if and only if $i>i'$ or $i=i'$ and $j>j'$. This gives a total order on the pairs $(i,j)$. In order to compute the coefficient of $x^m$ in $\frac{W_0(x)}{(1-x)^7}$, we have to consider finitely man cases, because each interval of integers separated by the numbers $ni+j/2$ can be treated uniformly.
For instance if $m=ni+j/2$, then the coefficient we are looking for is the sum of all $a_{i',j'}(n)\cdot\binom{n(i-i')+j/2-j'/2+6}{6}$ for all $(i',j')\le(i,j)$ (note that $1/(1-x)^7=\sum\binom{k+6}{6}x^k$).
In the other case, $ni+5<m<n(i+1)-2$. Then the coefficient is the sum of all $a_{i',j'}(n)\cdot\binom{m-ni'-j'/2+6}{6}$ for all $(i',j')\le(i,j)$.
Thus we obtain finitely many (in both cases together less than $300$) explicit polynomials in $n$ and $m$ (of degree $\le 14$ in $n$ and $\le 6$ in $m$) for which we have to prove that they are positive for all $n\ge2$ and all $m$ in the specified range.
I'm sure each of these steps is painful, but doable.