The closed form for $\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n$
We present here the details of calculating the closed form of the generating function.
$$s(z) = \sum_{n=1}^{\infty}\frac{z^n}{n^2} H_{n/2}\tag{1}$$
I have given partial results already in a comment.
In contrast to that of the OP in which a constant C appears the present calculation is complete.
We proceed step by step with the generating functions up to the quantity in question $g_{2}(z)$.
We shall do this with Mathematica taking care that theses two conditions are met
a) $g(z=0) = 0$
This is a necassary condition for the integration in the next step to be convergent at $0$.
b) $g(z)$ is real for $-1<z<1$
This almost always produces "nicer" expressions, i.e. they are better integrable in the next step than the "rough" expressions.
$$g_0(z) = \sum_{n=1}^{\infty}z^n H_{n/2}=\frac{z \log (4)+2 \log (1-z)}{z^2-1}\tag{2}$$
$$g_{1}(z) =\sum_{n=1}^{\infty}\frac{z^n}{n} H_{n/2} =\int_0^z \frac{g_0(t)}{t}\,dt\tag{3}$$
$$g_{2}(z) =\sum_{n=1}^{\infty}\frac{z^n}{n^2} H_{n/2} =\int_0^z \frac{g_1(t)}{t}\,dt\tag{4}$$
The indefinite integral using Integrate[]
related to $g_1(z)$ is
$$g_{1,i}(z) = \int \frac{g_0(z)}{z}\,dz=\operatorname{Li}_2\left(\frac{1-z}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \log ^2(1-z)+\log (z+1) \log (1-z)-\log (2) \log (z+1)$$
Subtracting the value at $z=0$ which is $\frac{1}{12} \left(\pi ^2-6 \log ^2(2)\right)$ gives for the definite integral $(3)$ the following expression
$$g_1(z) = \operatorname{Li}_2\left(\frac{1-z}{2}\right)+2 \operatorname{Li}_2(z)+\frac{1}{2} \log ^2(1-z)+\\ \log (z+1) \log (1-z)-\log (2) \log (z+1)+\frac{1}{12} \left(6 \log ^2(2)-\pi ^2\right)\tag{3a}$$
This expression meets the "nicety"- conditions requested.
Now the next step. The indefinite integral becomes
$$g_{2,i}(z) = \int \frac{g_1(z)}{z}\,dz=\text{expression with length 28}$$
Subtracting the value at $z=0$ which is $g_{2,i}(z=0) = -\frac{17 \zeta (3)}{8}-\frac{1}{6} \log ^3(2)$ gives an expression as the sum of 30 terms (in order to save typing labour (and errors), I have provided also the Mathematica expression in the apendix)
$$g_2(z) = \text{sum of 30 terms, see appendix}\tag{4a}$$
Here is the graph of $g_2$
Special values which have already been given in a comment are
$$g_2(z=+1) = \lim_{z\to 1^-} \, g_{2}(z)\\ = \frac{1}{4} \left(-4 \text{Li}_3(2)+9 \zeta (3)-2 i \pi \log ^2(2)+\pi ^2 \log (2)\right)= \frac{11}{8} \zeta (3)\tag{5}$$
$$\\g_2(z=-1) = \lim_{z\to -1^+} \, g_{2}(z)\\= \frac{1}{8} \left(-16 \text{Li}_3(2)+11 \zeta (3)-4 i \pi \log (2) \log (4)+\pi ^2 \log (16)\right)\\=-\frac{3}{8} \zeta (3)\tag{6}$$
Going from the immediate result of the limit to the final result we have used the transformation formulas for the polylog functions (see e.g. https://en.wikipedia.org/wiki/Polylogarithm).
Discussion
Splitting the sum into even and odd summands we have
$$g_2(z) =g_{2,e}(z)+g_{2,o}(z) $$
Since we have $g_2(z)$, and $g_{2,e}(z)$ is easily calculated with the result
$$g_{2,e}(z)=\frac{1}{4} \left(\operatorname{Li}_3\left(x^2\right)-\operatorname{Li}_3\left(1-x^2\right)+\operatorname{Li}_2\left(1-x^2\right) \log \left(1-x^2\right)\\ +\log (x) \log ^2\left(1-x^2\right)+\zeta (3)\right)\tag{7}$$
we have also obtained the more complicated sum
$$g_{2,o}(z) =\sum_{m=1}^{\infty} \frac{z^{2m-1}}{2m-1} H_{m-\frac{1}{2}} \\ =g_{2}(z)-g_{2,e}(z)\tag{8} $$
Appendix
Mathematica expression of $g_{2}(z)$
Notice that the transformation to a "nice", i.e. to an all-real summands-expression, still has to be done (my task):
g2[z]=Log[2]^3/6 - 1/12 \[Pi]^2 Log[z] + 1/2 Log[2]^2 Log[z] +
1/2 Log[1 - z]^2 Log[z] + Log[2] Log[z] Log[(2 z)/(1 + z)] +
1/2 (Log[(1 - z)/2] + Log[1/(1 + z)] -
Log[-((-1 + z)/(1 + z))]) Log[(2 z)/(1 + z)]^2 +
Log[(1 - z)/2] Log[z] Log[(1 + z)/2] -
1/2 Log[2] Log[z] (Log[4] + Log[z] - 2 Log[1 + z]) +
Log[1 - z] Log[z] Log[1 + z] +
1/2 (-Log[-z] + Log[z]) Log[
1 + z] (-2 Log[1 - z] + Log[1 + z]) + (Log[-z] - Log[z]) Log[
1 + z] Log[(1 + z)/(1 - z)] +
1/2 (Log[2/(1 - z)] + Log[z] - Log[-((2 z)/(1 - z))]) Log[(1 + z)/(
1 - z)]^2 +
Log[1 - z] PolyLog[2,
1 - z] + (Log[1 + z] - Log[(1 + z)/(1 - z)]) PolyLog[2, 1 - z] +
Log[z] PolyLog[2, 1/2 - z/2] +
Log[2] PolyLog[2, -z] + (Log[z/(1 + z)] + Log[1 + z]) PolyLog[2, z] +
Log[(2 z)/(
1 + z)] (PolyLog[2, z/(1 + z)] -
PolyLog[2, (2 z)/(1 + z)]) + (Log[z] -
Log[(2 z)/(1 + z)]) PolyLog[2, (1 + z)/
2] + (Log[1 - z] + Log[(1 + z)/(1 - z)]) PolyLog[2, 1 + z] +
Log[(1 + z)/(
1 - z)] (PolyLog[2, -((1 + z)/(1 - z))] -
PolyLog[2, (1 + z)/(1 - z)]) - 2 PolyLog[3, 1 - z] +
PolyLog[3, z] - PolyLog[3, z/(1 + z)] + PolyLog[3, (2 z)/(1 + z)] -
PolyLog[3, (1 + z)/2] - PolyLog[3, 1 + z] -
PolyLog[3, -((1 + z)/(1 - z))] + PolyLog[3, (1 + z)/(1 - z)] + (
17 Zeta[3])/8
Using $g_1(z)$ from @Dr. Wolfgang Hintze solution above
$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \ln ^2(1-z)+\ln (z+1) \ln (1-z)-\log (2) \ln (z+1)}$$
By algebraic identities, we have
$$\frac{1}{2} \log ^2(1-z)+\log (z+1) \log (1-z)=\frac{1}{2} \log ^2(1-z^2)-\frac{1}{2} \log ^2(1+z)$$
so
$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac{1}{2}\right)+2 \text{Li}_2(z)+\frac{1}{2} \ln ^2(1-z^2)-\frac{1}{2} \ln ^2(1+z)}-\ln(2)\ln(1+z)$$
And in this paper page $95$ Eq $(5)$ we have
$$\sum_{n=1}^\infty \overline{H}_n\frac{z^n}{n}=\operatorname{Li}_2\left(\frac{1-z}{2}\right)-\operatorname{Li}_2\left(\frac12\right)-\operatorname{Li}_2(-z)-\ln2\ln(1-z)$$
Subtracting the two generalizations we have
$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}z^n-\sum_{n=1}^\infty \frac{\overline{H}_n}{n}z^n=2 \text{Li}_2(z)+\text{Li}_2(-z)+\frac{1}{2} \ln ^2(1-z^2)-\frac{1}{2} \ln ^2(1+z)+\ln(2)\ln\left(\frac{1-z}{1+z}\right)}$$
Now divide both sides by $z$ then $\int_0^x$ we get
$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}x^n-\sum_{n=1}^\infty \frac{\overline{H}_n}{n^2}x^n$$ $$=2\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)+\frac12\underbrace{\int_0^x\frac{\ln^2(1-z^2)}{z}\ dz}_{\large I_1}-\frac12\underbrace{\int_0^x\frac{\ln^2(1+z)}{z}\ dz}_{\large I_2}$$ $$+\ln(2)[\operatorname{Li}_2(-x)-\operatorname{Li}_2(x)]$$
$I_1$ and $I_2$ can be found in the book Almost Impossible Integrals, Sums and Series page 3.
$$I_1=\int_0^x\frac{\ln^2(1-z^2)}{z}\ dz=\frac12\int_0^{x^2}\frac{\ln^2(1-t)}{t}\ dt$$
$$=\ln(x)\ln^2(1-x^2)+\ln(1-x^2)\operatorname{Li}_2(1-x^2)-\operatorname{Li}_3(1-x^2)+\zeta(3)$$
$$I_2=\ln(x)\ln^2(1+x)-\frac23\ln^3(1+x)-2\ln(1+x)\operatorname{Li}_2\left(\frac{1}{1+x}\right)-2\operatorname{Li}_3\left(\frac{1}{1+x}\right)+2\zeta(3)$$
$\sum_{n=1}^\infty \frac{\overline{H}_n}{n^2}x^n$ is already calculated here
$$\sum_{n=1}^\infty\frac{\overline{H}_{n}}{n^2}x^n=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)$$ $$-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3\left(\frac{1}{2}\right)+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]$$
Combine all results we get
$$\sum_{n=1}^\infty\frac{H_{n/2}}{n^2}x^n=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)+\operatorname{Li}_3\left(\frac{1}{1+x}\right)$$
$$-\frac12\operatorname{Li}_3(1-x^2)+\operatorname{Li}_3(x)+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]$$
$$+\frac12\ln(1-x^2)\operatorname{Li}_2(1-x^2)+\ln(1+x)\operatorname{Li}_2\left(\frac{1}{1+x}\right)+\ln(2)[\operatorname{Li}_2(-x)-\operatorname{Li}_2(x)]$$
$$-\frac12\ln(x)\ln^2(1+x)+\frac13\ln^3(1+x)+\frac12\ln(x)\ln^2(1-x^2)-\frac12\zeta(3)+\operatorname{Li}_3\left(\frac{1}{2}\right)$$