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.
Best Answer
First of all, thank you for the highly interesting question, it provides an excellent exercise!
Unfortunately it is not possible to improve the approximation by your method alone. Of course, the error in your approximation is due to taking the roots of $2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=e^{-\frac{3x}{2}}$ when $x\to\infty$ and thus using the asymptotic $2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=0$ instead. In order to alter this error, we examine your function $g(x)$:
$$ g(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+2)!}=\frac{e^{-x}\left(2e^{3x/2}\sin(\frac{1}{6}(\pi-3\sqrt3 x))-1\right)}{3x^2}, $$
and note that in order to change the properties of the sine wave on the RHS we need to alter the factorial on the LHS. Let $k$ now denote a positive integer $\geq0$ and examine what happens when we change $(3n+2)!$ to some $(3n+k)!$. In particular, we examine the three cases $k\equiv0 \ (\textrm{mod} \ 3)$, $k\equiv1 \ (\textrm{mod} \ 3)$ and $k\equiv2 \ (\textrm{mod} \ 3)$.
Case 1. $k=3j$, $j\in\mathbb{Z}$.
Let's examine $k=0$ first and generalize later.
$$ g'(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n)!}=-\frac13 e^{-x} \left(2e^{3x/2}\cos\left(\frac{\sqrt{3} x}{2}\right)+1\right), $$
and by similar methods we get $\zeta(3)\cong \frac87\left(\frac{\pi^{3}}{6(3)^{3/2}}\right)=1.136602043\ldots$ which is not a very good approximation. Having $j>0$ yields similar results that include the cosine function, but with some leading $\exp$ terms. For instance, we have for $j=3$:
$$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+9)!}=\frac{e^{-x}\left(-e^{x}x^6+120e^xx^3-720e^x+480e^{3x/2}\cos(\frac{\sqrt{3} x}{2})+240\right)}{720x^9} ,$$
and we can see that when $x\to\infty$ we are making an even bigger error in our approximation by assuming that its roots are those that satisfy the equation
$$ 480\cos\left(\frac{\sqrt{3} x}{2}\right)=0. $$
In fact, as $j$ increases so does our error.
Case 2. $k=3j+1$, $j\in\mathbb{Z}$.
In this case we get similar expressions as your original sine wave, but with some differences in sign (pun not intended). For $j=0$ we have
$$ g'(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+1)!}=-\frac{e^{-x}\left(2e^{3x/2}\sin(\frac{1}{6}(\pi+3\sqrt3 x))-1\right)}{3x^2}, $$
which has the same roots as your original function $g(x)$ and the same error in approximating $\zeta(3)$. As in Case 1, increasing the value of $j$ leads to even more leading $\exp$ terms and thus increases the error in our approximation.
Case 3. $k=3j+2$, $j\in\mathbb{Z}$.
The case $j=0$ corresponds to your original function $g(x)$. As with cases 1 and 2, increasing $j$ increases the amount of leading $\exp$ terms and thus the error in the approximation due to taking the roots of the asymptotic value when $x\to\infty$.
It remains to be seen whether it is possible to sensibly generalize the factorial to $(in+j)!$, $i,j\in\mathbb{Z}, i>0$ in order to get other approximations for other odd zeta values, perhaps several of these cases might yield good approximations. Euler himself tried to generalize his proof of the Basel problem to compute values for $\sum_n n^{-(2k+1)}$, $k>0$ but found no way to achieve this. Since then, considerable brain power has been expended to do just that, without success. I believe that is a good indicator that this path might not lead to a nice closed-form expression for $\zeta(3)$ in terms of $\pi^3$.
EDIT 11.12.2018
Taking the function
$$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{5n}}{(3n+2)!}=\frac{e^{-x^{5/3}}\left(2e^{3x^{5/3}/2}\sin(\frac{1}{6}(\pi-3\sqrt3 x^{5/3}))-1\right)}{3x^{10/3}} $$
yields the same approximation as $g(x)$, i.e., with $x^{3n}$ instead of $x^{5n}$ in the sum. This provides further indication that this approach is unlinkely to yield any better results on $\zeta(3)$.