The whole sum is:
$$ \sum _{n=1}^{\infty} \left( (-1)^n \frac{ \left(H_{n-1}-H_{\left\lfloor \frac{n-1}{2}\right\rfloor }\right)}{2n^3}+\frac{ \left(H_{n-1}\right){}^2-H_{n-1}^{(2)}}{2^{n+1} \: n^2}\right) = \frac{\pi^4}{120} -\frac{\ln^4(2)}{24} -\frac{7}{8}\ln(2)\zeta(3) \approx 0.07307223704$$
Where $H_n^{(k)} $ are the generalized harmonic numbers. The sum on the right is simply a Stirling Number sum, but the left is giving me trouble. My apologies if this has already been posted; I couldn't find something similar.
Using
$$\sum_{n=1}^\infty a_n=\sum_{n=0}^\infty a_{2n+1}+\sum_{n=1}^\infty a_{2n},$$ we have
$$S=\sum _{n=1}^{\infty} (-1)^n \frac{ H_{n-1}-H_{\left\lfloor \frac{n-1}{2}\right\rfloor }}{n^3}$$ $$=-\sum _{n=0}^{\infty} \frac{ H_{2n}-H_n}{(2n+1)^3}+\sum _{n=1}^{\infty}\frac{ H_{2n-1}-H_{n-1}}{(2n)^3}$$
$$=-\sum _{n=0}^{\infty} \frac{ H_{2n+1}-\frac{1}{2n+1}-H_n}{(2n+1)^3}+\sum _{n=1}^{\infty}\frac{ H_{2n}-\frac1{2n}-H_{n}+\frac{1}{n}}{(2n)^3}$$
$$=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}-\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^3}+\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac18\sum_{n=1}^\infty\frac{H_{n}}{n^3}$$
$$+\sum_{n=0}^\infty\frac{1}{(2n+1)^4}+\frac1{16}\sum_{n=1}^\infty\frac{1}{n^4}.$$
For the first two sums, use
$$\sum_{n=1}^\infty a_{2n}-\sum_{n=0}^\infty a_{2n+1}=\sum_{n=1}^\infty (-1)^n a_{n},$$
$$S=\sum_{n=1}^\infty\frac{(-1)^n H_{n}}{n^3}+\sum_{n=1}^\infty\frac{ H_{n}}{(2n+1)^3}+\frac18\sum_{n=1}^\infty\frac{ H_{n}}{n^3}+\sum_{n=0}^\infty\frac{1}{(2n+1)^4}+\frac1{16}\zeta(4).$$
We have: $$\sum_{n=1}^\infty\frac{(-1)^n H_{n}}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)+\frac{1}{12}\ln^4(2)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4);$$
which follows from
$$\sum_{n=1}^\infty\frac{H_n}{n^{2q+1}}=-\frac12\sum_{i=1}^{2q-1}(-1)^i\zeta(2q-i+1)\zeta(i+1)$$
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^{4}}=\frac{15}{16}\zeta(4);$$
which follows from
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^{a}}=(1-2^{-a})\zeta(a).$$
For the remaining sum,
$$\sum_{n=0}^\infty \frac{H_n}{(2n+1)^3}=\frac12\int_0^1 \ln^2(x)\sum_{n=0}^\infty H_n x^{2n}dx$$ $$=-\frac12\int_0^1\frac{\ln^2(x)\ln(1-x^2)}{1-x^2}dx=-\frac1{16}\int_0^1 \frac{x^{-1/2} \ln^2(x)\ln(1-x)}{1-x}dx$$
$$=-\frac1{16}\frac{d}{da}\int_0^1 \frac{x^{a-1} \ln(x)\ln(1-x)}{1-x}dx\bigg|_{a=1/2}$$
$$=-\frac1{16}\frac{d}{da} \left(\psi^{(1)}(a)[\psi(a)+\gamma]-\frac12\psi^{(2)}(a)\right)\bigg|_{a=1/2}$$ $$=-\frac1{16}\left(\left(\psi^{(1)}(a)\right)^2+\psi^{(2)}(a)[\psi(a)+\gamma]-\frac12\psi^{(3)}(a)\right)\bigg|_{a=1/2}$$
$$=\frac{45}{32}\zeta(4)-\frac{7}{4}\ln(2)\zeta(3).$$
Bonus: Using the same approach of evaluating $\sum_{n=1}^\infty \frac{H_n}{(2n+1)^3}$, we can get the generalization:
$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^q}=(2^{1-q}-2)\ln(2)\zeta(q)+q(1-2^{-q-1})\zeta(q+1)\nonumber\\ -\frac12\sum_{j=1}^{q-2}(2^{j+1}-1)(2^{-j}-2^{-q})\zeta(q-j)\zeta(j+1).$$