Following @mickep's idea, we can simplify the integral as follows:
1. Legendre chi function and its integral representation. Define
$$ \chi_2 (z) = \frac{\operatorname{Li}_2(z) - \operatorname{Li}_2(-z)}{2} = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)^2}. $$
We can check that (e.g. see my previous posting) the following integral representation holds
$$\int_{0}^{\frac{\pi}{2}} \arctan(z \sin\theta) \, d\theta = 2\chi_2\Big( \tfrac{\sqrt{1+z^2}-1}{z}\Big). \tag{$z \notin \pm i(1,\infty)$} $$
Moreover, we have the following two identities, which can be easily checked by differentiating both sides.
\begin{align*}
\chi_2(z) + \chi_2(\tfrac{1-z}{1+z}) &= \tfrac{1}{8}\pi^2 - \tfrac{1}{2}\log z \log(\tfrac{1-z}{1+z}), \\
\chi_2(z) + \chi_2(\tfrac{1}{z}) &= \tfrac{1}{4}\pi^2 + \tfrac{i}{2} \pi (\operatorname{sign}\Im z) \log z.
\end{align*}
2. Decomposition of arctangent. Using @mickep's idea, if we let $\omega = e^{2\pi i/3}$, then we have
\begin{align*}
\arctan(z^3)
&= \frac{\log(1+iz^3) - \log(1-iz^3)}{2} \\
&= -\sum_{k=0}^{2} \frac{\log(1+i\omega^k z) - \log(1-i\omega^k z)}{2} \\
&= -\sum_{k=0}^{2} \arctan(\omega^k z).
\end{align*}
3. Simplification of the integral using $\chi_2$. Combining two results, for $|z| \leq 1$ we have
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan(z^3 \sin^3\theta) \, d\theta
&= -2 \sum_{k=0}^{2} \chi_2\Big( \tfrac{\sqrt{1+\omega^{2k}z^k}-1}{\omega^k z}\Big).
\end{align*}
Now let $\alpha = \frac{-1+i}{2}(\sqrt{3} - 1)$. Then it is straightforward to check that
$$ \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3\theta) \, d\theta = -2 (\chi_2(\sqrt{2} - 1) + \chi_2(\alpha) + \chi_2(\bar{\alpha})) $$
and that $\frac{1-\alpha}{1+\alpha} = -\bar{\alpha}^{-1}$. Using the identities invloving $\chi_2$, we find that
$$ \chi_2(\alpha) + \chi_2(\bar{\alpha}) = -\tfrac{3}{32} \pi^2 + \tfrac{1}{8}\log^2(2+\sqrt{3}). $$
Combining altogether, we have
$$\int_{0}^{\frac{\pi}{2}} \arctan(\sin^3\theta) \, d\theta
= \tfrac{1}{16}\pi^2 + \tfrac{1}{2}\log^2(1+\sqrt{2}) - \tfrac{1}{4}\log^2(2+\sqrt{3}). $$
Addendum: a possible way of generalization. For odd $n = 2m+1$ and $\omega = e^{2\pi i/n}$, we may write
$$ \int_{0}^{\pi/2} \arctan(\sin^n t) \, dt
= (-1)^m \sum_{k=0}^{n-1} \int_{0}^{\pi/2} \arctan(\omega^k \sin t) \, dt. $$
Pairing up the conjugate terms and simplifying, this is written as the following form
$$ = (-1)^m 2 \chi_2(\sqrt{2}-1) + (-1)^m 2 \sum_{k=1}^{m} \epsilon_k (\chi_2(\alpha_k) + \chi_2(\bar{\alpha}_k)), $$
where $\epsilon_k \in \{-1, 1\}$ is appropriately chosen so that $\alpha_k$ is of the form
$$ \alpha_k = x_k + iy_k \quad \text{with } y_k = \sqrt{1+\smash[b]{2x_k - x_k^2}}. $$
Indeed, $(\epsilon_k)$ and $\alpha_k$ are chosen as follows:
$$ \epsilon_k = \begin{cases}
1 & \text{if } \Re(\omega^k) < 0, \\
-1 & \text{if } \Re(\omega^k) > 0,
\end{cases}
\quad\text{and} \quad
\alpha_k = \begin{cases}
\omega^{-k}(\sqrt{1+\omega^{2k}}-1) & \text{if } \Re(\omega^k) < 0, \\
-\omega^{k}(\sqrt{1+\omega^{-2k}}-1) & \text{if } \Re(\omega^k) > 0,
\end{cases}
$$
(In other words, we choose the sign $\epsilon_k$ so that $\epsilon_k \omega^k$ has always negative real parts.) Using the same idea as before, we can simplify
$$ \chi_2(\alpha_k) + \chi_2(\bar{\alpha}_k) = -\tfrac{1}{8}\pi^2 + \tfrac{1}{2}\left(\tfrac{\pi}{2} + \arctan\Big(\tfrac{x_k}{y_k}\Big) \right)^2 + \tfrac{1}{8}\log^2(1+2x_k). $$
For example, with aid of Mathematica, we can check that
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan(\sin^5 x) \, dx
&= \tfrac{1}{8}\pi^2 - \tfrac{1}{2}\log^2\left(1+\sqrt{2}\right) \\
&\quad - \tfrac{1}{4} \log^2 \left(\tfrac{1}{2} \left(1+\sqrt{5}-\sqrt{2 (1+\smash[b]{\sqrt{5}})}\right)\right) \\
&\quad + \tfrac{1}{4} \log^2 \left(\tfrac{1}{2} \left(3+\sqrt{5}-\sqrt{10+6 \smash[b]{\sqrt{5}}}\right)\right) \\
&\quad +\left(\pi -\tan^{-1}\left(\sqrt{2+\smash[b]{\sqrt{5}}}\right) - \cot^{-1}(5^{1/4}) \right) \\
&\qquad \times \left(\cot^{-1}(5^{1/4}) - \tan^{-1}\left(\sqrt{2+\smash[b]{\sqrt{5}}}\right)\right)
\end{align*}