Based on Three-sided coin's comments, it seems
$$I_4 = \int_0^1 \frac{\operatorname{Li}_4(x)}{1+x}dx =\ln(2)\zeta(4)+\tfrac34\zeta(2)\zeta(3)-\tfrac{59}{32}\zeta(5) \approx 0.321352\dots$$
(Note: Confirmed as correct by Brevan's answer.) Connecting it to Three-sided coin's other integrals, then
$$I_4 = \ln(2)\zeta(4)-\tfrac12\zeta(2)\zeta(3)-h_1$$
$$I_4 = \ln(2)\zeta(4)+\tfrac14\zeta(2)\zeta(3)-h_2$$
where,
$$h_1=\int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(-x)}}{x}dx = -\tfrac54\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
$$h_2=\int_0^1\frac{\color{blue}{\operatorname{Li}_3(-x)}\ln(1-x)}{x}dx = -\tfrac12\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
which implies,
$$h_1-h_2 =\int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(-x)}{x}dx- \int_0^1\frac{\operatorname{Li}_3(-x)\ln(1-x)}{x}dx =-\tfrac34\zeta(2)\zeta(3)$$
Compare to the similar integrals here that he mentioned,
$$h_3= \int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(x)}}{x}dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
$$h_4 = \int_0^1\frac{\color{blue}{\operatorname{Li}_3(x)}\ln(1-x)}{x}dx =\zeta(2)\zeta(3)-3\zeta(5$$
which has the proven relation,
$$h_3-h_4 = \int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(x)}{x}dx - \int_0^1\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}dx = \zeta(2)\zeta(3)$$
Update: Per Brevan's answer:
$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx=(-1)^nA(n)+\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)$$
with Dirichlet eta function $\eta(k)$ and
$$A(n) = \sum_{k=1}^\infty \frac{H_k}{k^n}(-1)^k$$
is the $n$th "Alternating Euler Sum". However, since,
$$A(n,z) = \sum_{k=1}^\infty \frac{H_k}{k^n}z^k =S_{n-1,2}(z)+\operatorname{Li}_{n+1}(z)$$
for $-1\leq z\leq 1$, then $I_n$ can be expressed by Nielsen polylogs $S_{n,p}(z)$ as I suspected.
Partial answer:
$$\begin{align}
I_1&=\int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\,dx\\
&\stackrel{x\to\sqrt{1-t^2}}{=}
\int_{0}^1 \cosh(\sqrt{1-t^2})\sin(t)\frac{t\,dt}{\sqrt{1-t^2}}\\
&=-\int_{0}^1 \sin(t)\,d(\sinh(\sqrt{1-t^2}))\\
&=\left[-\sin(t)\sinh(\sqrt{1-t^2})\right]_0^1+\int_{0}^1 \sinh(\sqrt{1-t^2})\,d (\sin(t))\\
&=\int_{0}^1 \sinh(\sqrt{1-t^2})\cos(t)\, dt=I_2.
\end{align}$$
It follows $I_1=I_2=\frac\pi4$.
Applying the same method one can show $I_3+I_4=\cosh(1)-\cos(1)$.
Best Answer
Integrating by parts 3 times,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= - \frac{x^{4}}{3} \cot(x) \left(\csc^{2} (x) +2 \right) \Bigg|^{\pi/2}_{0} + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \csc^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{2}{3}x^{3} \cot^{2}(x) \Bigg|^{\pi/2}_{0} + 2 \int_{0}^{\pi /2} x^{2} \cot^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + 2 \int_{0}^{\pi /2} x^{2} \cot^{2} (x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx -2x^{2} \Big( x + \cot(x) \Big) \Bigg|^{\pi/2}_{0} +4 \int_{0}^{\pi /2} x\Big(x+ \cot(x) \Big) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{4} + 4 \int_{0}^{\pi /2} x^{2} \ dx + 4 \int_{0}^{\pi /2} x \cot(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{12} + 4 \int_{0}^{\pi /2} x \cot(x) \ dx . \end{align}$$
In general, $$ \int_{a}^{b} f(x) \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{a}^{b} f(x) \sin (2nx) \ dx .$$
See here.
So $$ \begin{align} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x^{3} \sin (2nx) \ dx \\ &= 2 \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1} \pi^{3}}{16n} - \frac{(-1)^{n-1} 3\pi}{8n^{3}} \right) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{3 \pi}{4} \eta(3) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3). \end{align}$$
And
$$ \begin{align} \int^{\pi /2}_{0} x \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x \sin(2nx) \ dx \\ &= -\frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \\ &= \frac{\pi \ln 2}{2} . \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= \frac{8}{3} \left( \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3) \right) - \frac{\pi^{3}}{12} + 4 \left(\frac{\pi \ln 2}{2} \right) \\ &= - \frac{\pi^{3}}{12} + 2 \pi \ln(2) + \frac{\pi^{3}}{3} \ln (2) - \frac{3 \pi}{2} \zeta(3) . \end{align}$$