$$x_n=\frac1n-\frac1{n+1}+\frac1{n+2}-...=\sum_{k=0}^\infty \frac{(-1)^k}{n+k}=\int_0^1\frac{x^{n-1}}{1+x}dx$$
$$\Longrightarrow x_n^2=\int_0^1\int_0^1\frac{(xy)^{n-1}}{(1+x)(1+y)}dxdy$$
$$=\int_0^1\int_0^y\frac{t^{n-1}}{(y+t)(1+y)}dxdy$$
$$=\int_0^1t^{n-1}\left(\int_t^1\frac{dy}{(y+t)(1+y)}\right)dt$$
$$=\int_0^1\, t^{n-1}\frac{2\ln(\frac{1+t}{2})-\ln(t)}{1-t}dt$$
$$\Longrightarrow\sum_{n=1}^\infty\frac{1}{n^3}x_n^2=\int_0^1\, \frac{2\ln(\frac{1+t}{2})-\ln(t)}{1-t}\left(\sum_{n=1}^\infty\frac{t^{n-1}}{n^3}\right)dt$$
$$=\int_0^1\, \frac{2\ln(\frac{1+t}{2})-\ln(t)}{1-t}\left(\frac{\operatorname{Li}_3(t)}{t}\right)dt$$
$$=\int_0^1\, \frac{2\ln(\frac{1+t}{2})-\ln(t)}{t}\operatorname{Li}_3(t)dt+\underbrace{\int_0^1\, \frac{2\ln(\frac{1+t}{2})-\ln(t)}{1-t}\operatorname{Li}_3(t)dt}_{IBP}$$
$$=\int_0^1\, \frac{2\ln(\frac{1+t}{2})-\ln(t)}{t}\operatorname{Li}_3(t)dt+\int_0^1\frac{\operatorname{Li}_2(1-t)\operatorname{Li}_2(t)}{t}dt-2\int_0^1\frac{\operatorname{Li}_2(\frac{1-t}{2})\operatorname{Li}_2(t)}{t}dt$$
$$=I_1+I_2-2I_3$$
$$I_1=2\int_0^1\frac{\ln(1+t)\operatorname{Li}_3(t)}{t}dt-2\ln(2)\int_0^1\frac{\operatorname{Li}_3(t)}{t}dt-\int_0^1\frac{\ln(t)\operatorname{Li}_3(t)}{t}dt$$
$$=-2\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 t^{n-1}\operatorname{Li}_3(t)dt-2\ln(2)\zeta(4)+\zeta(5)$$
$$=-2\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)-2\ln(2)\zeta(4)+\zeta(5)$$
$$=\frac{75}{16}\zeta(5)-2\ln(2)\zeta(4)-\frac32\zeta(2)\zeta(3)$$
$$I_2=\int_0^1\frac{\operatorname{Li}_2(1-t)\operatorname{Li}_2(t)}{t}dt$$
$$=\int_0^1\frac{(\zeta(2)-\ln(t)\ln(1-t)-\operatorname{Li}_2(t))\operatorname{Li}_2(t)}{t}dt$$
$$=\zeta(2)\zeta(3)-\underbrace{\int_0^1\frac{\ln(t)\ln(1-t)\operatorname{Li}_2(t)}{t}dt}_{IBP}-\int_0^1\frac{\operatorname{Li}_2^2(t)}{t}dt$$
$$=\zeta(2)\zeta(3)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(t)}{t}dt$$
$$=\frac92\zeta(5)-2\zeta(2)\zeta(3).$$
I was not able to find $I_3$ but I will try again.
Different approach with a full solution:
$$x_n=\frac1n-\frac1{n+1}+\frac1{n+2}-...=\sum_{k=0}^\infty \frac{(-1)^k}{n+k}=(-1)^n\left(\overline{H}_n+\frac{(-1)^n}{n}-\ln(2)\right)$$
$$\Longrightarrow x_n^2=\overline{H}_n^2+2\frac{(-1)^n\overline{H}_n}{n}-2\ln(2)\overline{H}_n-2\ln(2)\frac{(-1)^n}{n}+\frac1{n^2}+\ln^2(2)$$
$$\Longrightarrow \sum_{n=1}^\infty\frac{1}{n^3}x_n^2=\sum_{n=1}^\infty\frac{\overline{H}_n^2}{n^3}+2\sum_{n=1}^\infty\frac{(-1)^n\overline{H}_n}{n^4}-2\ln(2)\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$$
$$+\frac74\ln(2)\zeta(4)+\zeta(5)+\ln^2(2)\zeta(3)\tag{1}$$
The second and third sums are known:
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\frac74\ln(2)\zeta(3)-\frac5{16}\zeta(4)\tag{2}$$
$$\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{4}}=\frac{59}{32}\zeta(5)-\frac{15}{8}\ln(2)\zeta(4)-\frac{3}{4}\zeta(2)\zeta(3)\tag{3}$$
For the first one, we use
$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_{2n}+\sum_{n=0}^\infty a_{2n+1}$$
$$\Longrightarrow \sum_{n=1}^\infty\frac{\overline{H}_n^2}{n^3}=\sum_{n=1}^\infty\frac{\overline{H}_{2n}^2}{(2n)^3}+\sum_{n=0}^\infty\frac{\overline{H}_{2n+1}^2}{(2n+1)^3}$$
$$=\sum_{n=1}^\infty\frac{(H_{2n}-H_n)^2}{(2n)^3}+\sum_{n=0}^\infty\frac{(H_{2n+1}-H_n)^2}{(2n+1)^3}$$
$$=\color{blue}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^3}}-2\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n)^3}+\sum_{n=1}^\infty\frac{H_{n}^2}{(2n)^3}+\color{blue}{\sum_{n=0}^\infty\frac{H_{2n+1}^2}{(2n+1)^3}}$$
$$\color{red}{-2\sum_{n=0}^\infty\frac{H_nH_{2n+1}}{(2n+1)^3}}+\sum_{n=0}^\infty\frac{H_{n}^2}{(2n+1)^3}$$
$$=\color{blue}{\sum_{n=1}^\infty\frac{H_n^2}{n^3}}-2\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n)^3}+\sum_{n=1}^\infty\frac{H_{n}^2}{(2n)^3}$$
$$\color{red}{-2\sum_{n=0}^\infty\frac{H_nH_{2n}}{(2n+1)^3}-2\sum_{n=0}^\infty\frac{H_n}{(2n+1)^4}}+\sum_{n=0}^\infty\frac{H_{n}^2}{(2n+1)^3}$$
$$=-2\sum_{n=0}^\infty\frac{H_nH_{2n}}{(2n+1)^3}-2\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n)^3}$$
$$+\sum_{n=0}^\infty\frac{H_{n}^2}{(2n+1)^3}-2\sum_{n=0}^\infty\frac{H_n}{(2n+1)^4}+\frac98\sum_{n=1}^\infty\frac{H_n^2}{n^3}$$
The first two sums are evaluated by Cornel here:
$$\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln^3(2)\zeta (2) -\frac{7}{8} \ln^2(2)\zeta (3)$$
$$-\frac{1}{15} \ln^5(2)-2 \ln(2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)$$
$$\sum _{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n+1)^3}=\frac{1}{12}\ln^5(2)+\frac{31}{128} \zeta (5)-\frac{1}{2} \ln^3(2)\zeta (2)+\frac{7}{4} \ln^2(2) \zeta (3)$$
$$-\frac{17}{8} \ln(2)\zeta (4)+2\ln(2) \operatorname{Li}_4\left(\frac{1}{2}\right).$$
The third sum is given here
$$\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}=\frac{31}{8}\zeta(5)-\frac{45}{8}\ln2\zeta(4)+\frac72\ln^22\zeta(3)-\frac78\zeta(2)\zeta(3)$$
The last two are well known:
$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}{8}\ln(2)\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)$$
$$\sum_{n=1}^\infty \frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$$
Collecting these results, we get
$$\sum _{n=1}^{\infty } \frac{\overline{H}_n^2}{n^3}=4\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac{167}{32}\zeta(5)+\frac{19}{8}\ln(2)\zeta(4)+\frac74\ln^2(2)\zeta(3)$$
$$+\frac13\ln^3(2)\zeta(2)+\frac34\zeta(2)\zeta(3)-\frac1{30}\ln^5(2)\tag{4}$$
Finally, plug $(2), (3),$ and $(4)$ in $(1)$,
$$\sum_{n=1}^\infty\frac{1}{n^3}x_n^2=4\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac{17}{32}\zeta(5)+\ln(2)\zeta(4)-\frac34\ln^2(2)\zeta(3)$$
$$+\frac13\ln^3(2)\zeta(2)-\frac34\zeta(2)\zeta(3)-\frac1{30}\ln^5(2)=0.4940843..$$
