The only other real numbers with this property are $\pm 1/n!$. (Note that these are indeed \emph{other} values than $\pm 1, \dots, \pm n$ because $n>1$.) Define the polynomial $q(x) = x^{2n+2}-x^{2n}p(1/x) = x^{2n+2}-(a_0x^{2n}+\cdots+a_{2n-1}x+1)$. The statement that $p(1/x)=x^2$ is equivalent (for $x\neq 0$) to the statement that $x$ is a root of $q(x)$. Thus we know that $\pm 1,\pm 2,\ldots,\pm n$ are roots of $q(x)$, and we can write $$q(x) = (x^2+ax+b)(x^2-1)(x^2-4)\cdots (x^2-n^2)$$ for some monic quadratic polynomial $x^2+ax+b$. Equating the coefficients of $x^{2n+1}$ and $x^0$ on both sides gives $0=a$ and $-1=(-1)^n(n!)^2 b$, respectively. Since $n$ is even, we have $x^2+ax+b = x^2-(n!)^{-2}$. We conclude that there are precisely two other real numbers $x$ such that $p(1/x)=x^2$, and they are $\pm 1/n!$.