The following (hopefully) settles the question. 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 $n\ge2$. We show that *all*
coefficients of $H(x)$ are positive whenever $n\ge6$.

I'm not sure about the easiest kind to prove the assertion about the coefficients of $H(x)$. But the following works: 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. Write $H(x)=H_0(x^2)+xH_1(x^2)$ and $W(x)=W_0(x^2)+xW_1(x^2)$. Then $H_0(x)(1-x)^7=W_0(x)$ and $H_1(x)(1-x)^7=W_1(x)$. Thus we are reduced to show that $H_0(x)$ and $H_1(x)$ have nonnegative coefficients. In the following we write $W_\star$ and $H_\star$ for $W_0$ and $W_1$, and $H_0$ and $H_1$, respectively. We obtain these coefficients by multiplying $W_\star(x)$ with the formal power series $\frac{1}{(1-x)^7}=\sum\binom{k+6}{6}x^k$.

All what follows is assisted/confirmed by the Sage code below.

There is a finite set of pairs $(i,j)$ of nonnegative integers, *independent* of $n$, such that the exponents of $W_\star(x)$ have the form $ni+j$. Let $a_{i,j}$ be the corresponding coefficient. It is a polynomial in $n$. It turns out that $0\le i\le 11$ and $-2\le j\le 5$.

We have
\begin{equation} 
W_\star = \sum_{i,j}a_{i,j}(n)x^{ni+j}.
\end{equation}
(Of course the $a_{i,j}$ and the pairs $(i,j)$ are different for $W_0$ and $W_1$.)

If $n\ge8$, then $ni+j>ni'+j'$ if and only if $i>i'$ or $i=i'$ and $j>j'$. This gives a total order on the pairs $(i,j)$. Every exponent of $x$ in $W_\star(x)/(1-x)^7$ which matters has the form $ni+j+y$ where $0\le y<n(i'-i)+j'-j$, where $(i',j')$ is the successor of $(i,j)$ for our order.

The coefficient of $x^{ni+j+y}$ in $W_\star(x)\frac{1}{(1-x)^7}$
is the sum of all $a_{i_1,j_1}(n)\cdot\binom{ni+j-ni_1-j_1+y+6}{6}$ for all $(i_1,j_1)\le(i,j)$.

Thus we obtain *finitely* many (actually $126$) explicit polynomials $f(n,y)$ in $n$ and $y$ (of small degrees in $n$ and $y$) for which we have to prove that they are positive for all $n\ge8$ and all $y$ in the specified range.

We distinguish two cases: If $(n'i+j')-(ni+j)=1$ (recall that $(i',j')$ is the successor of $(i,j)$ again), then only $y=0$ is possible. In all these cases, we see that $f(n+6,0)$ has nonnegative coefficients. Thus $f(n,0)\ge0$ whenever $n\ge6$.

Now assume that $(n'i+j')-(ni+j)>1$. In these cases it turns out that $i'=i+1$, so $0\le y< n-e$ for some integer $e$. Thus $y=n-e-1-o$ for some non-negative $o$. Thus $f(n,o)=f(y+e+1+o,o)=g(y,o)$. This latter polynomial, in $y$ and $o$, has to be non-negative for all non-negative $o$ and $y$. It turns out that except for a single example actually oll the coefficients are nonnegative. For this single exception, $g(y,o+1)$ has nonnegative coefficients. Thus $g(y,o)\ge0$ except possibly for $o=0$. But $g(y,0)$ has nonnegative coefficients, and things are settled.

Our argument used $n\ge8$. For $n=6$ and $7$, one can directly compute $H(x)$ and check that the coefficients are nonnegative. The cases $2\le n\le5$ are checked directly for $V(x)$.

*Remark (answering Jason's question):* It is a known fact that a polynomial $V$ is positive on $(0,1)$ if and only if it is a nonnegative linear combination of polynomials $x^i(1-x)^j$. The problem is that it may involve terms where $i+j$ is bigger than $\deg V$. (This doesn't happen here, though.) I used an LP solver to play a little with $V$ and $\frac{V}{(1-x^2)^4x^{2n-2}}$. From that a pattern showed up which lead to a solution.

By the way, $V$ actually seems to be a nonnegative linear combination of $x^i(1-x)^{\deg V-i}$. This can be obtained from the recursion (fixed $n$)
\begin{equation}
V_0=V,\;\; V_{i+1}=\frac{V_i-V_i(1)x^{\deg V_i}}{1-x}.
\end{equation}
I have an explicit expression of the coefficients $a_i$ in $V=\sum a_ix^i(1-x)^{\deg V-i}$. But I don't see an easy argument showing positivity.

The scrollable <pre> field where I had put the Sage code cuts off part of the program. (An MO bug?) Instead you find it [here][1] now.


  [1]: https://www.mathematik.uni-wuerzburg.de/~mueller/amde_proof.sage