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 + 32
\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 whenever $n\ge6$. 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 $ni+j/2$ (left bound including, right bound excluding) can be treated uniformly.

More concretely: Let $(i_1,j_1)<(i_2,j_2)$ be consecutive pairs, that is either $i_1=i_2$ and $j_2=j_1+1$, or $i_2=i_1+1$, $j_1=5$, $j_2=-2$. Then if $ni_1+j_1/2\le m<ni_2+j_2/2$, then the coefficient we are looking for is the sum of all $a_{i,j}(n)\cdot\binom{m-ni-j/2+6}{6}$ for all $(i,j)\le(i_1,j_1)$. (Note that $1/(1-x)^7=\sum\binom{k+6}{6}x^k$.)

Thus we obtain *finitely* many (namely $128$) 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 $n\ge6$ and all $m$ in the specified range.

**Added:** Along these lines, I now have a complete rigorous proof. The positivity of these finitely many polynomials is actually quite easy to prove, it essentially happens because they have (under suitable substitutions) positive coefficients. Later today I hope to have the time to write up the details and provide the cleaned up Sage code which proves the claim.