What follows in an evaluation of the Euler sum that uses its equivalent integral representation.
We begin by noting that
$$\int_0^1 x^{2n} \ln^2 x \, dx = \frac{d^2}{ds^2} \left [\int_0^1 x^{2n + s} \, dx \right ]_{s = 0} = \frac{2}{(2n + 1)^3}.$$
Thus
$$\sum_{n = 1}^\infty \frac{(-1)^{n} H_n}{(2n + 1)^3} = \frac{1}{2} \int_0^1 \ln^2 x \sum_{n = 1}^\infty (-1)^n H_n x^{2n} \, dx\tag1$$
From the Generating function for the Harmonic numbers, namely
$$\sum_{n = 1}^\infty H_n x^n = -\frac{\ln (1 - x)}{1 - x},$$
enforcing a substitution of $x \mapsto -x^2$ one has
$$\sum_{n = 1}^\infty (-1)^n H_n x^{2n} = -\frac{\ln (1 + x^2)}{1 + x^2},$$
allowing us to rewrite (1) as
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = -\frac{1}{2} \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx.$$
Evaluating the integral we have
\begin{align}
I &= \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx\\
&= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - \underbrace{\int_1^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx}_{\displaystyle x \mapsto 1/x}\\
&= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx + 2 \int_0^1 \frac{\ln^3 x}{1 + x^2} \, dx \tag2\\
&= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 12 \beta (4) - I\\
2 I &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 12 \beta (4).
\end{align}
Note the right most integral in (2) was found as follows:
$$\int_0^1 \frac{\ln^3 x}{1 + x^2} \, dx = \sum_{n = 0}^\infty (-1)^n \frac{d^3}{ds^3} \left [\int_0^1 x^{2n + s} \, dx \right ]_{s = 0} = -6 \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^4} = -6 \beta (4).$$
Here $\beta (x)$ is the Dirichlet beta function and has a known value in terms of the polygamma function of order 3 of:
$$\beta (4) = \frac{1}{768} \left [\psi^{(3)} \left (\frac{1}{4} \right ) - 8 \pi^4 \right ].$$
Thus
$$I = \frac{1}{2}\int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 6 \beta (4).$$
To find the last outstanding integral, set $x = \tan \theta$, then
\begin{align}
I_1 &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx\\
&= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\tan \theta) \ln (\cos \theta) \, d\theta\\
&= - 2 \int_0^{\frac{\pi}{2}} \Big{(} \ln (\sin \theta) - \ln (\cos \theta) \Big{)}^2 \ln (\cos \theta) \, d\theta\\
&= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\sin \theta) \ln (\cos \theta) \, d\theta + 4 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln^2 (\cos \theta) \, d\theta - 2 \int_0^{\frac{\pi}{2}} \ln^3 (\cos \theta) \, d\theta\\
&= I_\alpha + I_\beta + I_\gamma.
\end{align}
Each of the above three integrals can be found by taking third derivatives of the Beta function.
For $I_\alpha$
\begin{align}
I_\alpha &= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\sin \theta) \ln (\cos \theta) \, d\theta\\
&= -\frac{1}{8} \lim_{x,y \to 1/2} \frac{\partial^3}{\partial x^2 \partial y} \operatorname{B} (x,y)\\
&= -\frac{1}{8} \left (2 \pi \zeta (3) - 8\pi \ln^3 2 \right )\\
&= -\frac{\pi}{4} \zeta (3) + \pi \ln^3 2
\end{align}
For $I_\beta$
\begin{align}
I_\beta &= 4 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln^2 (\cos \theta) \, d\theta\\
&= \frac{1}{4} \lim_{x,y \to 1/2} \frac{\partial^3}{\partial x \partial y^2} \operatorname{B} (x,y)\\
&= \frac{1}{4} \left (2 \pi \zeta (3) - 8\pi \ln^3 2 \right )\\
&= \frac{\pi}{2} \zeta (3) - 2\pi \ln^3 2
\end{align}
For $I_\gamma$
\begin{align}
I_\gamma &= -2 \int_0^{\frac{\pi}{2}} \ln^3 (\cos \theta) \, d\theta\\
&= -\frac{1}{8} \lim_{y \to 1/2} \frac{\partial^3}{\partial y^3} \operatorname{B} \left (\frac{1}{2},y \right )\\
&= -\frac{1}{8} \left (-12 \pi \zeta (3) - 8\pi \ln^3 2 -2 \pi^3 \ln 2\right )\\
&= -\frac{3\pi}{2} \zeta (3) + \pi \ln^3 2 + \frac{\pi^3}{4} \ln 2.
\end{align}
Thus
$$I_1 = \frac{7}{4} \pi \zeta (3) + \frac{\pi^3}{4} \ln 2,$$
giving
$$I = \frac{7}{8} \pi \zeta (3) + \frac{\pi^3}{8} \ln 2 - 6 \beta (4),$$
which finally leads to the following value for the Euler sum of
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = 3 \beta (4) - \frac{7}{16} \pi \zeta (3) - \frac{\pi^3}{16} \ln 2,$$
or
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = \frac{1}{256} \psi^{(3)} \left (\frac{1}{4} \right ) - \frac{\pi^4}{32} - \frac{7 \pi}{16} \zeta (3) - \frac{\pi^3}{16} \ln 2,$$
as claimed.