We shall use $A_n=\sum _{k=1}^n \frac{(-1)^{k+1}}{k}$ (and for completeness $A_0 = 0$) for the alternating harmonic sum.
The sum in question is
$$s = \sum_{k\ge 0} (-1)^k (A_k-\log(2))^2\\= \sum_{k\ge 0} \left((A_{2k}-\log(2))^2-(A_{2k+1}-\log(2))^2\right)$$
The summand can be expanded to
$$A_{2 k}^2-A_{2 k+1}^2-2 \log (2) A_{2 k}+2 \log (2) A_{2 k+1} \\
= -\frac{2 (A_{2 k}-\log (2))}{2 k+1}-\frac{1}{(2 k+1)^2}\tag{1}$$
Here we have used that for integer $k$
$$A_{k+1}-A_{k}=\frac{(-1)^k}{k+1}\tag{2}$$
So we have
$$s = s_1 + s_2$$
with
$$s_1 =2 \sum_{k\ge 0} \frac{ \log(2) - A_{2 k}}{2 k+1}\tag{3}$$
and
$$s_2 = -\sum_{k\ge0} \frac{1}{(2 k+1)^2}= -\frac{\pi^2}{8}\tag{4}$$
which we have summed up imediately.
In the first one we use that for integer $k$
$$A_{2 k}-\log(2) = \int_0^1 \frac{1-x^{2 k}}{x+1} \, dx-\int_0^1 \frac{1}{x+1} \, dx\\
= -\int_0^1 \frac{x^{2 k}}{x+1} \, dx\tag{5}$$
so that the sum under the $x$-integral becomes
$$ \sum_{k\ge0} \frac{2 x^{2 k} }{2 k+1}=\frac{2 \operatorname{arctanh}(x)}{x}=\frac{1}{x} \log(\frac{1+x}{1-x})\tag{6}$$
Hence the first sum is given by the integral
$$s_1 = \int_0^1 \frac{1}{x(1+x)} \log(\frac{1+x}{1-x})=\frac{\pi^2}{6}\tag{7}$$
Finally we get
$$s = s_1 + s_2 =\frac{\pi^2}{6} -\frac{\pi^2}{8}= \frac{\pi^2}{24}$$
Q.E.D.
Discussion
1) Integral representations of the alternating and odd harmonic numbers
Inserting $\frac{1}{k}=\int_0^1 x^{k-1}\,dx$ into the definition of $A_n$ and doing the sum under the integral we obtain
$$A_n=\int_0^1 \frac{1-(-1)^n x^n}{x+1}\,dx\tag{8a}$$
for $n=2k$ this simplifies to
$$A_{2k}=\int_0^1 \frac{1- x^{2k}}{x+1}\,dx\tag{8b}$$
as used in $(5)$.
Similarly we have for the odd harmonic number the representation
$$O_n = \int_0^1 \frac{x^{2 n}-1}{x^2-1} \, dx\tag{9}$$
2) The introduction of an integral in $(4)$ is a deviation from the requirement of the OP to use only series operations.
Ideally we would like to see $s_1= \sum_{k\ge 1} \frac{1}{k^2}$
Hence my "solution" is incomplete. I'll try to improve it.
EDIT
I'm almost there.
I find by series manipulations (changing the order of summation) that
$$s_1 = \sum_{n\ge1} \frac{O_{n}}{(n+1)(2n-1)}\tag{10}$$
Where the odd harmonic sum is defined as
$$O_n=\sum_{m=1}^n \frac{1}{2m-1}\tag{11}$$
It is easy to show that
$$O_n = H_{2n} -\frac{1}{2} H_{n}\tag{12}$$
so that
$$s_1 = \sum_{n\ge1} \frac{H_{2n} -\frac{1}{2} H_{n}}{(n+1)(2n-1)}\tag{13}$$
But we have (https://en.wikipedia.org/wiki/Harmonic_number)
$$H_{2z}=\log(2) + \frac{1}{2}(H_z+H_{z-\frac{1}{2}})\tag{14}$$
so that
$$s_1 = \sum_{n\ge1} \frac{\log(2) -\frac{1}{2} H_{n-\frac{1}{2}}}{(n+1)(2n-1)}\tag{15}$$
The first sum yields $2 \log(2)^2$. The second sum
$$s_{1b} =- \frac{1}{2}\sum_{n\ge1} \frac{ H_{n-\frac{1}{2}}}{(n+1)(2n-1)}= ?\tag{16}$$
Still has to be calculated.
Summarizing: the numerical result of $s_1$ agrees with $\pi^2/6$,
Best Answer
Another approach
Using the same strategy of @omegadot,
from this paper page $105$ we have
$$\overline{H}_n=\ln2-\int_0^1\frac{(-x)^n}{1+x}\ dx$$
multiply both sides by $\frac{(-1)^nH_n}{n^2}$ then $\sum_{n=1}^\infty$ we get
$$S=\ln2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\underbrace{\int_0^1\frac{1}{1+x}\sum_{n=1}^\infty\frac{H_nx^n}{n^2}\ dx}_{\large \mathcal{I}}\tag1$$
From here we have
$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
$$\Longrightarrow \mathcal{I}=\underbrace{\int_0^1\frac{\operatorname{Li}_3(x)}{1+x}\ dx}_{\large \mathcal{I}_1}-\underbrace{\int_0^1\frac{\operatorname{Li}_3(1-x)}{1+x}\ dx}_{\large \mathcal{I}_2}+\underbrace{\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{1+x}\ dx}_{\large \mathcal{I}_3}$$ $$+\underbrace{\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx}_{\large \mathcal{I}_4}+\zeta(3)\underbrace{\int_0^1\frac{1}{1+x}\ dx}_{\ln2}$$
$$\mathcal{I}_1=\int_0^1\frac{\operatorname{Li}_3(x)}{1+x}\ dx=-\sum_{n=1}^\infty(-1)^n\int_0^1 x^{n-1}\operatorname{Li}_3(x)\ dx$$ $$=-\sum_{n=1}^\infty(-1)^n\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)$$
$$=\ln2\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$
$$\mathcal{I}_2=\int_0^1\frac{\operatorname{Li}_3(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\int_0^1\frac{\operatorname{Li}_3(x)}{2-x}\ dx$$ $$=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\operatorname{Li}_3(x)\ dx =\sum_{n=1}^\infty\frac1{2^n}\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)$$
$$=\ln2\zeta(3)-\zeta(2)\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n}{2^nn^3}$$
$$\mathcal{I}_3=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{2-x}\ dx$$
$$=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln x\operatorname{Li}_2(x) \ dx=\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2H_n}{n^3}+\frac{H_n^{(2)}}{n^2}-\frac{2\zeta(2)}{n^2}\right)$$
$$=2\sum_{n=1}^\infty\frac{H_n}{2^nn^3}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^nn^2}-2\zeta(2)\operatorname{Li}_2\left(\frac12\right)$$
$$\mathcal{I}_4=\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\frac12\int_0^1\frac{\ln(1-x)\ln^2x}{2-x}\ dx$$
$$=\frac12\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln(1-x)\ln^2x \ dx$$ $$=\frac12\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2\zeta(3)}{n}+\frac{2\zeta(2)}{n^2}-\frac{2H_n}{n^3}-\frac{2H_n^{(2)}}{n^2}-\frac{2H_n^{(3)}}{n}\right)$$
$$=\ln2\zeta(3)+\zeta(2)\operatorname{Li}_2\left(\frac12\right)-\sum_{n=1}^\infty\frac{H_n}{2^nn^3}-\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^nn^2}-\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}$$
Combine the results of $\mathcal{I}_1$, $\mathcal{I}_2$, $\mathcal{I}_3$ and $\mathcal{I}_4$
$$\Longrightarrow \mathcal{I}=2\ln2\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}$$
now plug this result in $(1)$
$$ \Longrightarrow S=\frac54\zeta(4)-2\ln2\zeta(3)+\ln2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$
Finally, substitute
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac58\zeta(3)\tag{i}$$
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}=\operatorname{Li}_4\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42\tag{ii}$$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li}_4\left(\frac12\right)-\frac{11}{4}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42\tag{iii}$$
we obtain
Note that the results of $(i)$ and $(ii)$ follow from using the the generating functions
$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}x^n=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x).$$
As for $(iii)$, its already calculated here.
The interesting thing about this approach is that some tough series got cancelled and we used only well-known results of harmonic series.