Define $g_n(x)=\sqrt{1-\sqrt{1-\cdots\sqrt{1-\sqrt{x}}}}$, where there are $n$ radicals present. For instance $g_2(x)=\sqrt{1-\sqrt{x}}$ and $g_1(x) = \sqrt{x}$. Notice that for $n\ge 2$, $g_n(x)=g_{n-1}(1-\sqrt{x})$. Also define $$ I_{n,m} = \int_0^1 x^m \sin^{-1}(g_n(x))dx \overset{1-\sqrt{x}=t}{=} 2\int_0^1 (1-t)^{2m+1}\sin^{-1}(g_{n-1}(t))dt. $$ A recurrence relation is formed and we obtain $$ I_{n,m} = 2\sum_{r=0}^{2m+1}\binom{2m+1}{r}(-1)^rI_{n-1,r},$$ where we keep reducing until we reach terms of the form $I_{1,l}$ for $l\in \mathbb{N}$. This integral can be computed exactly, $$ I_{1,m} = \int_0^1 x^m \sin^{-1}(\sqrt{x})dx = 2\int_0^1 t^{2m+1}\sin^{-1}(t)dt = \frac{\pi}{2(m+1)}\left(1-\frac{\Gamma(m+3/2)}{\sqrt{\pi} (m+1)!} \right)$$ This also shows that the integral $I_{n,0}$ is always of the form $\frac{p}{q}\pi$ where $p$ and $q$ are rational numbers.
Addendum : An explanation of how I got the final integral. Using integration by parts and this result, $$\begin{align*}\int_0^1 t^{2m+1}\sin^{-1}(t)dt &= \left.\sin^{-1}(t) \frac{t^{2m+2}}{2m+2}\right|_{0}^1 -\frac{1}{2(m+1)}\int_0^1 \frac{t^{2m+2}}{\sqrt{1-t^2}}dt\\ &= \frac{\pi}{4(m+1)} - \frac{1}{2(m+1)} \cdot \sqrt{\pi}\frac{\Gamma(m+3/2)}{2\Gamma(m+2)} \\ &= \frac{\pi}{4(m+1)}\left(1-\frac{\Gamma(m+3/2)}{\sqrt{\pi} (m+1)!} \right).\end{align*}$$ For additional clarity on how the integrals are calculated, $$I_{1,0} = \frac{\pi}{2}\left[1-\frac{\Gamma\left(\frac{3}{2}\right)}{\sqrt{\pi}\cdot 1!}\right] = \frac{\pi}{2}\left[1-\frac{1}{2}\right] = \frac{\pi}{4}\\ I_{1,1} = \frac{\pi}{4}\left[1-\frac{\Gamma\left(\frac{5}{2}\right)}{\sqrt{\pi} \cdot 2!}\right] = \frac{\pi}{2}\left[1-\frac{3}{8}\right] = \frac{5\pi}{32} \\ I_{1,2}= \frac{\pi}{6}\left[1-\frac{\Gamma\left(\frac{7}{2}\right)}{\sqrt{\pi} \cdot 3!}\right] = \frac{\pi}{2}\left[1-\frac{5}{16}\right] = \frac{11\pi}{96} \\ I_{1,3}= \frac{\pi}{8}\left[1-\frac{\Gamma\left(\frac{9}{2}\right)}{\sqrt{\pi} \cdot 4!}\right] = \frac{\pi}{2}\left[1-\frac{35}{128}\right] = \frac{93\pi}{1024} $$ From these four integrals, $$I_{2,0} = 2I_{1,0}-2I_{1,1} = 2\cdot \frac{\pi}{4}-2\cdot \frac{5\pi}{32} = \frac{3}{16}\pi.$$ and $$I_{3,0} = 2I_{2,0}-2I_{2,1} = \frac{3\pi}{8}-4\left[I_{1,0}-3I_{1,1}+3I_{1,2}-I_{1,3}\right] \\ = \frac{3\pi}{8} - \pi +\frac{\pi}{2}+\frac{93}{256}\pi = \frac{61\pi}{256}$$ You have a lot of patience if you have computed $I_{4,0}$ by hand.
Addendum 2: Since you have asked for $I_{n,0}$, using the recurrence relation, $I_{n,0} = 2(I_{n-1,0}-I_{n-1,1}) = 2(2(I_{n-2,0}-I_{n-2,1})-2(I_{n-2,0}-3I_{n-2,1}+3I_{n-2,2}-I_{n-2,3}))$, thus, $$ \boxed{I_{n,0} = 4\big[ 2I_{n-2,1}-3I_{n-2,2}+I_{n-2,3}\big]}$$