It seems to have escaped attention that these sums may be evaluated
using harmonic summation techniques.
Introduce the sum
$$S(x; \alpha, p) = \sum_{n\ge 1} \frac{1}{n^p(e^{\alpha n x}-1)}$$
with $p$ a positive odd integer and $\alpha>1$, so that we seek e.g.
$2 S(1; \pi\sqrt{2}, 3)+S(1; 2\pi\sqrt{2}, 3).$
The sum term is harmonic and may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^p}, \quad \mu_k = k
\quad \text{and} \quad
g(x) = \frac{1}{e^{\alpha x}-1}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{1}{e^{\alpha x}-1} x^{s-1} dx
= \int_0^\infty \frac{e^{-\alpha x}}{1-e^{-\alpha x}} x^{s-1} dx
\\ = \int_0^\infty \sum_{q\ge 1} e^{-\alpha q x} x^{s-1} dx
= \sum_{q\ge 1} \int_0^\infty e^{-\alpha q x} x^{s-1} dx
\\= \Gamma(s) \sum_{q\ge 1} \frac{1}{(\alpha q)^s}
= \frac{1}{\alpha^s} \Gamma(s) \zeta(s).$$
It follows that the Mellin transform $Q(s)$ of the harmonic sum
$S(x;\alpha,p)$ is given by
$$Q(s) = \frac{1}{\alpha^s} \Gamma(s) \zeta(s) \zeta(s+p)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} =
\sum_{k\ge 1} \frac{1}{k^p} \frac{1}{k^s}
= \zeta(s+p)$$
for $\Re(s) > 1-p.$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the left for an expansion about
zero.
First formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(2 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+3).$$
We shift the Mellin inversion integral to the line $s=-1$, integrating
right through the pole at $s=-1$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^3\sqrt{2}}{72 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(3)$$
and
$$\frac{1}{2}\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^3\sqrt{2}x}{36}.$$
This almost concludes the proof of the first formula if we can show
that the integral on the line $\Re(s) = -1$ vanishes when $x=1.$ To
accomplish this we must show that the integrand is odd on this line.
Put $s = -1 - it$ in the integrand to get
$$\pi^{1+it} \sqrt{2}^{1+it}
\left(2 + 2^{1+it}\right)
\Gamma(-1-it) \zeta(-1-it) \zeta(2-it).$$
Now use the functional equation of the Riemann Zeta function in the
following form:
$$\zeta(1-s) = \frac{2}{2^s\pi^s}
\cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$
to obtain (with $s=-1-it$)
$$\pi^{1+it} \sqrt{2}^{1+it}
\left(2 + 2^{1+it}\right)
\zeta(2+it) 2^{-1-it} \pi^{-1-it}
\frac{1}{2\cos\left(\frac{\pi (-1-it)}{2}\right)}
\zeta(2-it)$$
which is
$$ \sqrt{2}^{1+it}
\left(2^{-it} + 1\right)
\zeta(2+it)
\frac{1}{2\cos\left(\frac{\pi (1+it)}{2}\right)}
\zeta(2-it)$$
and finally yields
$$-\frac{1}{\sin(\pi i t/2)}
\left(\sqrt{2}^{-1-it}+\sqrt{2}^{-1+it}\right)
\zeta(2+it)\zeta(2-it).$$
It is now possible to conclude by inspection: the zeta function terms
and the powers of the square root are even in $t$ and the sine term is
odd, so the whole term is odd and the integral vanishes. (We get exponential decay from the sine term.)
Second formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(4 - \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+5).$$
We shift the Mellin inversion integral to the line $s=-2$ (no pole on
the line this time) picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^5\sqrt{2}}{540 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(5)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^5\sqrt{2}x}{540}.$$
It remains to verify that the integrand on the line $\Re(s)=-2$ is odd
when $x=1$. Put $s=-2-it$ in the integrand to get
$$\pi^{2+it} \sqrt{2}^{2+it}
\left(4 - 2^{2+it}\right)
\Gamma(-2-it) \zeta(-2-it) \zeta(3-it).$$
Applying the functional equation once again with $s=-2-it$ we obtain
$$\pi^{2+it} \sqrt{2}^{2+it}
\left(4 - 2^{2+it}\right)
\zeta(3+it) 2^{-2-it} \pi^{-2-it}
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is
$$\sqrt{2}^{2+it}
\left(2^{-it} - 1\right)
\zeta(3+it)
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is in turn
$$\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right)
\frac{1}{2\cos\left(\frac{\pi(2+it)}{2}\right)}
\zeta(3+it)
\zeta(3-it)$$
which finally yields
$$-\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right)
\frac{1}{2\cos(\pi i t /2)}
\zeta(3+it)
\zeta(3-it)$$
The product of the zeta function terms is even, as is the cosine
term. The term in front is odd, so the integrand is odd as claimed.
(We get exponential decay from the cosine term.)
Third formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(8 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+7).$$
We shift the Mellin inversion integral to the line $s=-3$, integrating
right through the pole at $s=-3$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{17 \pi^7\sqrt{2}}{37800 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{9}{2}\zeta(7)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^7\sqrt{2}x}{1134}
\quad\text{and}\quad
\frac{1}{2} \mathrm{Res}(Q(s)/x^s; s=-3) =
-\frac{\pi^7\sqrt{2}x^3}{4050}.$$
This almost concludes the proof of this third formula if we can show
that the integral on the line $\Re(s) = -3$ vanishes when $x=1.$ To
accomplish this we must show once more that the integrand is odd on
this line.
Put $s = -3 - it$ in the integrand to get
$$\pi^{3+it} \sqrt{2}^{3+it}
\left(8 + 2^{3+it}\right)
\Gamma(-3-it) \zeta(-3-it) \zeta(4-it).$$
By the functional equation we obtain with $s = -3-it$
$$\pi^{3+it} \sqrt{2}^{3+it}
\left(8 + 2^{3+it}\right)
\zeta(4+it)
2^{-3-it} \pi^{-3-it}
\frac{1}{2\cos\left(\pi(-3-it)/2\right)}
\zeta(4-it)$$
which is
$$\sqrt{2}^{3+it}
\left(2^{-it} + 1\right)
\zeta(4+it)
\frac{1}{2\cos\left(\pi(3+it)/2\right)}
\zeta(4-it)$$
which finally yields
$$\frac{1}{2\sin(\pi it/2)}
\left(\sqrt{2}^{3-it} + \sqrt{2}^{3+it}\right)
\zeta(4+it)
\zeta(4-it).$$
This concludes it since the two zeta function terms together are even
as is the square root term while the sine term is odd, so their
product is odd.
A similar yet not quite the same computation can be found at this MSE link.
Another computation in the same spirit is at this MSE link II.
I bet this can be considered as "cheating", but for any $a>0$ we have
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+a+1)^3(n+a+2)} = \frac{1+a}{2a^3}+\frac{1}{2a^3(a+1)}+\frac{1}{2}\psi''(a)\tag{1} $$
by partial fraction decomposition, and $\psi''(1)=\zeta(3), \psi''\left(\frac{1}{2}\right)=-14\,\zeta(3)$, $\psi''\left(\frac{3}{2}\right)=16-14\,\zeta(3), \psi''\left(\frac{5}{2}\right)=16+\frac{16}{27}-14\,\zeta(3)$ and
$$ \psi''\left(m+\frac{1}{2}\right)=16\left[\sum_{k=1}^{m}\frac{1}{(2k-1)^3}-\frac{7}{8}\zeta(3)\right]\tag{2} $$
By choosing a rather large value of $a$ like $a=\frac{7}{2}$, by $(1)$ and $(2)$ we get
$$ 7\zeta(3) = \frac{9741838}{1157625}-\sum_{n\geq 0}\frac{1}{\left(n+\frac{7}{2}\right)\left(n+\frac{9}{2}\right)^3\left(n+\frac{11}{2}\right)}\tag{3} $$
where the involved series has a positive value, but less than $10^{-3}$, leading to an accurate approximation for $\zeta(3)$ of the wanted type. Similarly
$$2\cdot 10^{-4}\approx\sum_{n\geq 0}\frac{1}{\left(n+\frac{11}{2}\right)\left(n+\frac{13}{2}\right)^3\left(n+\frac{15}{2}\right)}=\frac{4550782178678}{540820405125}-7\zeta(3)\tag{4}$$
and so on. By considering the continued fraction of $\frac{4550782178678}{3785742835875}$ and truncating it at the fourth convergent, we get $\zeta(3)\approx\color{red}{\frac{113}{94}}$ with an approximation error that is less than $10^{-4}$. $\frac{6}{5}$ is just the approximation we get by stopping at the third convergent.
It is interesting to point out that $\zeta(3)\approx\frac{6}{5}$ arises from the computation of the average order of the arithmetic function $\sigma(n)^2$. Since by Euler's product
$$ \forall s>3,\qquad \sum_{n\geq 1}\frac{\sigma(n)^2}{n^s}=\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)} $$
by tauberian theorems we have $\sum_{n\leq x}\sigma(n)^2 = \frac{5\zeta(3)}{6}x^3+O(x^2)$. On the other hand $3\zeta(4)\approx\zeta(3)\zeta(2)^2$ leads to $\zeta(3)\approx\frac{6}{5}$ and $\sum_{n\leq x}\sigma(n)^2\approx n^3$.
Best Answer
$$\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}x^k}{k^2\binom{2k}{k}}=2\,\text{arcsinh}^2\left(\frac{x}{2}\right)\tag{1}$$ due to the Maclaurin series of the squared arcsine, hence the computation of the series appearing in your conjecture is equivalent to the computation of the following integral: $$ I = \int_{0}^{1}\frac{2}{x}\text{arcsinh}^2\left(\frac{x}{2}\right)\log^2(x)\,dx.\tag{2} $$ In 1981 Leschiner pointed out a nice consequence of creative telescoping: $$\begin{eqnarray*}\sum_{n\geq 0}\left(1-\frac{1}{2^{2n+1}}\right)a^{2n+2}\zeta(2n+2)&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2-a^2}\\&=&\frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}\cdot\frac{3k^2+a^2}{k^2-a^2}\prod_{m=1}^{k-1}\left(1-\frac{a^2}{m^2}\right)\tag{3}\end{eqnarray*}$$ and in 2006 Bailey, Borwein and Bradley pointed out that similarly $$\begin{eqnarray*}\sum_{n\geq 0}\zeta(2n+2)a^{2n}&=&\sum_{k\geq 1}\frac{1}{k^2-a^2}\\&=&3\sum_{k\geq 1}\frac{1}{\binom{2k}{k}(k^2-a^2)}\prod_{m=1}^{k-1}\frac{m^2-4a^2}{m^2-a^2}\tag{4}\end{eqnarray*} $$ holds. By comparing the coefficients of $a^0$ in the LHS and RHS of $(3)$ we get the well-known $$ \frac{1}{2}\zeta(2)=\frac{3}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}.$$ The coefficient of $a^4$ in the LHS of $(3)$ is $$ \frac{7}{8}\zeta(4) = \frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}\left[\frac{4}{k^2}-3 H_{k-1}^{(2)}\right]$$ and the series $$ \sum_{k\geq 1}\frac{H_{k-1}^{(2)}}{k^2\binom{2k}{k}} $$ can be computed from $(4)$ or from the Maclaurin series of $\arcsin^4(x)$ (identity $(20)$ here), proving $$ \sum_{k\geq 1}\frac{\color{red}{1}}{k^4 \binom{2k}{k}} = \frac{17}{36}\zeta(4).\tag{5}$$