jmc

Let $q(x)=(x^2-1)p(x)-x.$ Then $q(x)$ has degree 7, and $q(n)=0$ for $n=2$, 3, 4, $\dots ,$ 7, so $$q(x)=(ax+b)(x-2)(x-3)\cdots (x-7)$$for some constants $a$ and $b.$

We know that $q(1)=(1^2-1)p(1)-1=-1.$ Setting $x=1$ in the equation above, we get $$q(1)=720(a+b),$$so $a+b=-\frac{1}{720}.$

We also know that $q(-1)=((-1{)}^2-1)p(-1)+1=1.$ Setting $x=-1$ in the equation above, we get $$q(-1)=20160(-a+b),$$so $-a+b=\frac{1}{20160}.$ Solving for $a$ and $b,$ we find $a=-\frac{29}{40320}$ and $b=-\frac{3}{4480}.$ Hence, $$\begin{array}{rl}q(x)& =\left(-\frac{29}{40320}x-\frac{3}{4480}\right)(x-2)(x-3)\cdots (x-7)\\ & =-\frac{(29x+27)(x-2)(x-3)\cdots (x-7)}{40320}.\end{array}$$In particular, $$q(8)=-\frac{(29\cdot 8+27)(6)(5)\cdots (1)}{40320}=-\frac{37}{8},$$so $$p(8)=\frac{q(8)+8}{8^2-1}=\boxed{\frac{3}{56}}.$$