jmc

Let $q(x)=xp(x)-1.$ Then $q(x)$ has degree 7, and $q(2^n)=0$ for $n=0,$ 1, 2, $\dots ,$ 6, so $$q(x)=c(x-1)(x-2)(x-2^2)\cdots (x-2^6)$$for some constant $c.$

We know that $q(0)=0\cdot p(0)-1.$ Setting $x=0$ in the equation above, we get $$q(0)=c(-1)(-2)(-2^2)\cdots (-2^6)=-2^{21}c,$$so $c=\frac{1}{2^{21}}.$ Hence, $$\begin{array}{rl}q(x)& =\frac{(x-1)(x-2)(x-2^2)\cdots (x-2^6)}{2^{21}}\\ & =(x-1)\left(\frac{x}{2}-1\right)\left(\frac{x}{2^2}-1\right)\cdots \left(\frac{x}{2^6}-1\right).\end{array}$$The coefficient of $x$ in $q(x)$ is then $$\begin{array}{rl}& [(1)(-1)(-1)\cdots (-1)]+\left[(-1)\left(\frac{1}{2}\right)(-1)\cdots (-1)\right]+\left[(-1)(-1)\left(\frac{1}{2^2}\right)\cdots (-1)\right]+\left[(-1)\cdots (-1)\left(-\frac{1}{2^6}\right)\right]\\ & =1+\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^6}=\frac{1-\frac{1}{2^7}}{1-\frac{1}{2}}=2-\frac{1}{64}=\frac{127}{64}.\end{array}$$Also, the constant coefficient in $q(x)$ is $-1,$ so $q(x)$ is of the form $$q(x)=\frac{1}{2^{21}}{x}^7+\cdots +\frac{127}{64}x-1.$$Then $$p(x)=\frac{q(x)+1}{x}=\frac{1}{2^{21}}{x}^6+\cdots +\frac{127}{64}.$$Therefore, $p(0)=\boxed{\frac{127}{64}}.$