Here is a partial attempt. Perhaps this may lead to a closed form.
Integrating by parts twice yields
$$
\begin{align}
\int_0^\infty\sin^{2n}(t)e^{-t}\,\mathrm{d}t
&=2n\int_0^\infty\sin^{2n-1}(t)\cos(t)\,e^{-t}\,\mathrm{d}t\\
&=2n\int_0^\infty\left((2n-1)\sin^{2n-2}(t)-2n\sin^{2n}(t)\right)\,e^{-t}\,\mathrm{d}t\\
(4n^2+1)\int_0^\infty \sin^{2n}(t)e^{-t}\,\mathrm{d}t
&=(4n^2-2n)\int_0^\infty\sin^{2n-2}(t)\,e^{-t}\,\mathrm{d}t
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\sin^{2n}(t)e^{-t}\,\mathrm{d}t
&=\prod_{k=1}^n\frac{2k(2k-1)}{4k^2+1}\\
&=\frac{(2n)!}{\prod\limits_{k=1}^n(4k^2+1)}
\end{align}
$$
Now
$$
\begin{align}
\int_0^\infty e^{-t}\log(\cos^2(t))\,\mathrm{d}t
&=\int_0^\infty\log\left(1-\sin^2(t)\right)\,e^{-t}\,\mathrm{d}t\\
&=-\sum_{n=1}^\infty\int_0^\infty\frac1n\sin^{2n}(t)\,e^{-t}\,\mathrm{d}t\\
&=-2\sum_{n=1}^\infty\frac{(2n-1)!}{\prod\limits_{k=1}^n(4k^2+1)}
\end{align}
$$
The terms of the last sum decay like $n^{-3/2}$, so the sum does converge, although slowly.
As derived in my other post, the integral is equal to the simpler looking sum
$$
\int_0^\infty e^{-t}\log(\cos^2(t))\,\mathrm{d}t=\sum_{k=1}^\infty(-1)^k\frac{8k}{4k^2+1}
$$
At least $4k^2+1$ is still in the denominator.