Infinite sums that include zeros of functions, that we cannot compute in closed-form, can sometimes be evaluated in closed-form with help of the residue theorem. A good reference for this is Bak and Newman's "Complex Analysis" (see e.g. Chapter 19.1 for a related example and Chapter 13.2 for the general theory). The hard part is finding the exact function to integrate. After some trial and error I ended up with
$$f(z) = \frac{\sin(z) - \cos^2(z)\sinh(z)}{z^6\cos(z)(\cos(z)\cosh(z)+1)}.$$
We basically want to start with the summand, add $g(z) = \cos(z)\cosh(z)+1$ in the denominator and $g'(z)$ in the numerator (to make the residues come out correct) since all the poles are simple. We can then massage the function as we like while keeping the residue fixed (e.g. making replacements like $\cosh(z)\to -1/\cos(z)$ since this holds at the poles of interest) and making sure the other residues are calculable and that we can find a suitable contour to integrate over where we have analytical control over the integral.
Let's start by computing the poles and the residues of our function $f$. Since $g(z)=g(-z)=g(iz)$ these are located on both the real axis and the imaginary axis (i.e. if we let $x_n$ be the positive real poles then the others are $-x_n$,$ix_n$ and $-ix_n$). The residue at each of these poles is (by construction) given by
$$\text{Res}(f\mid \cos(z)\cosh(z)+1=0) = \left(\frac{\sin(z)-\sinh(z)}{\cos(z)+\cosh(z)}\right)^2\frac{1}{z^6}.$$
This is an even function and for the poles on the imaginary axis the residues are seen to be equal the corresponding residues on the real axis. Thus the total contribution from all these poles are $4$ times the sum of the positive real ones.
The residue at the pole at $z=0$ is found (tedious calculation done with Mathematica) to be
$$\text{Res}(f\mid z=0) = \frac{1}{12}.$$
The residues at the poles of the cosine, $z_n = \pi(n+1/2)$ with $n\in\mathbb{Z}$, are
$$\text{Res}(f\mid z = \pi(n+1/2)) = -\frac{1}{\pi^6(n+1/2)^6},$$
and the sum of these residues is given by the sum quoted in the question: $\sum_{n\in\mathbb{Z}}\frac{1}{\pi^6(n+1/2)^6} = \frac{2}{15}$ (see also this).
At this point we notice that if we sum all the residues of all the poles in the complex plane and demand that this sum has to be zero then we arrive at
$$\boxed{4\sum_{n=1}^\infty \left(\frac{\sin(x_n)-\sinh(x_n)}{\cos(x_n)+\cosh(x_n)}\right)^2\frac{1}{x_n^6} = -\frac{1}{12} + \sum_{n\in\mathbb{Z}}\frac{1}{\pi^6(n+1/2)^6} = \frac{1}{20}},$$
which is the desired result.
To complete the derivation we have to find a series of contours $\gamma_{R_n}$ that for $n\to\infty$ encloses all the poles and for which we can show that $$\lim_{n\to\infty}\oint_{\gamma_{R_n}} f(z) dz = 0.$$
Once this is established the result then follows from the residue theorem.
The natural first choice is to try integrating the function over a square centered at the origin and make sure the sides are well separated from the poles by using a sidelength $R_n = 4\pi n$ (e.g. the right vertical edge will then be an integral over $z = 2\pi n + iy$ with $y$ ranging from $-2n\pi\to 2n\pi$. The integral over each of these horizonal/vertical sides are equal since $f(z)=-f(-z)$ so we only need to estimate one horizontal and one vertical integral.
Along the edge $z = 2n\pi + iy$ with $y\in[-2\pi n,2\pi n]$ we have
$$|f(2n\pi + iy)| = \frac{|\tanh(y) - \cosh(y)\sin(y)|}{|2n\pi+iy|^6|\cosh(y)\cosh(n\pi + iy)+1|} < \frac{8}{|2n\pi|^6}\text{ for } n\gg 1.$$
and the same bound is found to apply for $|f(x+2\pi n i)|$. A standard ML estimate then gives us $$\oint_{\gamma_{R_n}} f(z)dz \ll 4\cdot \frac{8}{(2\pi n)^6}\cdot (4\pi n) \to 0 \text{ for } n\to\infty.$$
