There is a rather neat proof of this.
First, note that there is already an analogue for this:
DLMF §10.21 says that a Rayleigh
function $\sigma_n(\nu)$ is defined as a similar power series
$$ \sigma_n(\nu) = \sum_{m\geq1} y_{\nu, m}^{-n}. $$
It links to http://arxiv.org/abs/math/9910128v1 among others as an example of how
to evaluate such things.
In your case, call $\zeta_m = y_{\nu,m}$ and $z=y_{\nu-1,k}$ ($\nu$ is $n$ shifted by $1$), so that after
expanding in partial fractions your sum is
$$ \sum_{m\geq1} \frac{\zeta_m z}{(\zeta_m-z)^3} = \sum_{m\geq1}
\frac{z^2}{(\zeta_m-z)^3} + \frac{z}{(\zeta_m-z)^2}. $$
Introduce the function
$$ y_\nu(z) = z^{-\nu/2}J_\nu(z^{1/2}). $$
By DLMF 10.6.5 its derivative
satisfies the two relations
$$\begin{aligned}
y'_\nu(z) &= (2z)^{-1} y_{\nu-1}(z) - \nu z^{-1} y_\nu(z)
\\&=
-\tfrac12 y_{\nu+1}(z).
\end{aligned} $$
It also has the infinite product
expansion
$$ y_\nu(z) = \frac{1}{2^\nu\nu!}\prod_{k\geq1}(1 - z/\zeta_k). $$
Therefore, each partial sum of $(\zeta_k-z)^{-s}$, $s\geq1$ can be evaluated in
terms of derivatives of $y_\nu$:
$$ \sum_{k\geq1}(\zeta_k-z)^{-s} = \frac{-1}{(s-1)!}\frac{d^s}{dz^s}\log
y_\nu(z). $$
When evaluating this logarithmic derivative, the derivative $y'_\nu$
can be expressed in terms of $y_{\nu-1}$, going down in $\nu$, but the derivative
$y'_{\nu-1}$ can be expressed in terms of $y_\nu$ using the other
relation that goes up in the index $\nu$. So even higher-order derivatives contain only $y_\nu$ and $y_{\nu-1}$.
I calculated your sum using this procedure with a CAS as:
$$ -\tfrac12z^2(\log y)''' -z(\log y)''
= \tfrac18\nu + z^{-1} P\big(y_{\nu-1}(z)/y_\nu(z)\big), $$
where $P$ is the polynomial
$$ P(q) = -\tfrac18 q^3 + (\tfrac38\nu-\tfrac18) q^2 + (-\tfrac14\nu^2
+ \tfrac14\nu - \tfrac18)q. $$
When $z$ is chosen to be any root of $y_{\nu-1}$,
$z=\mathsf{BesselJZero}[\nu-1, k]\hat{}2$, $P(q)=0$, your sum is equal
to
$$ \frac{\nu}{8}, $$
which is $(n+1)/8$ in your notation.
It is possible to derive a number of such closed forms for sums of
this type. For example, by differentiating $\log y$ differently
(going $\nu\to\nu+1\to\nu$), one would get
$$ \sum_{m\geq1}
\frac{y_{\nu,m}y_{\nu+1,k}}{(y_{\nu,m}-y_{\nu+1,k})^3} =
-\frac{\nu}{8}. $$
Some other examples, for which the r.h.s. is independent of $z$ ($\zeta_m=y_{\nu,m}, z=y_{\nu-1,l}$, $l$ arbitrary):
$$ \begin{gathered}
\sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^2} = \frac14,\\
\sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^4} - \frac{1}{(\zeta_k-z)^2} + \frac1{24}\frac{5-\nu}{\zeta_k-z} = \frac{1}{48}, \\
\sum_{k\geq1} \frac{\zeta_k}{(\zeta_k-z)^4} + \frac1{96}\frac{z-\zeta_k-8+4\nu}{(\zeta_k-z)^2} = 0.
\end{gathered} $$
or with $z=y_{\nu+1,l}$, $l$ arbitrary:
$$ \begin{gathered}
\sum_{k\geq1} \frac{z^2}{(\zeta_k-z)^3} = -\tfrac18\nu-\tfrac14,
\end{gathered} $$
and they get messier with higher degrees.
