We seek a closed form of
$$\sum_{p=0}^{n-1} \sum_{q=0}^n
|n-p-q| {n+p-q\choose p} {n-p+q-1\choose q}.$$
First part
We get for the argument to the absolute value being positive the
contribution
$$\sum_{p=0}^{n-1} \sum_{q=0}^{n-1-p}
(n-p-q) {n+p-q\choose p} {n-p+q-1\choose q}.$$
This is
$$\sum_{p=0}^{n-1} \sum_{q=0}^{n-1-p}
(q+1) {n+p-(n-1-p-q)\choose p}
{n-p+(n-1-p-q)-1\choose n-1-p-q}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^{n-1-p}
(q+1) {2p+q+1\choose p}
{2n-2p-q-2\choose n-p-q-1}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^p
(q+1) {2(n-1-p)+q+1\choose n-1-p}
{2n-2(n-1-p)-q-2\choose n-(n-1-p)-q-1}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^p
(q+1) {2n-2p+q-1\choose n-1-p}
{2p-q\choose p-q}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{p=0}^{n-1} z^p (1+z)^{-2p}
\sum_{q=0}^p
(q+1) (1+z)^q
{2p-q\choose p-q}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{p=0}^{n-1} z^p (1+z)^{-2p} [w^p] (1+w)^{2p}
\\ \times \sum_{q=0}^p
(q+1) (1+z)^q w^q (1+w)^{-q}.$$
We may extend the inner sum to infinity due to the coefficient
extractor in $w$, getting
$$[z^{n-1}] (1+z)^{2n-1}
\sum_{p=0}^{n-1} z^p (1+z)^{-2p} [w^p] (1+w)^{2p}
\frac{1}{(1-(1+z)w/(1+w))^2}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{p=0}^{n-1} z^p (1+z)^{-2p} [w^p] (1+w)^{2p+2}
\frac{1}{(1+w-(1+z)w)^2}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{p\ge 0} z^p (1+z)^{-2p} [w^p] (1+w)^{2p+2}
\frac{1}{(1-wz)^2}.$$
Here we have extended to infinity due to the extractor in $z.$
The contribution from $w$ is
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{p+1}} (1+w)^{2p+2} \frac{1}{(1-wz)^2}.$$
Now put $w/(1+w)=v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$ to
get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{p+1}} \frac{1}{(1-v)^{p+1}}
\frac{1}{(1-vz/(1-v))^2} \frac{1}{(1-v)^2}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{p+1}} \frac{1}{(1-v)^{p+1}}
\frac{1}{(1-v-vz)^2}.$$
Next put $v = (1-\sqrt{1-4u})/2$ so that $v(1-v)=u$ and $dv =
1/\sqrt{1-4u} \; du$ to obtain
$$\;\underset{u}{\mathrm{res}}\;
\frac{1}{u^{p+1}}
\frac{1}{(1-(1+z)(1-\sqrt{1-4u})/2)^2}
\frac{1}{\sqrt{1-4u}}.$$
Evaluate at $u=z/(1+z)^2$ to cancel the remaining sum (substitution
rule), here we get
$$\sqrt{1-4u} = \sqrt{1-4z/(1+z)^2}
= \frac{1}{1+z} \sqrt{(1+z)^2-4z}
= \frac{1-z}{1+z}$$
and obtain
$$[z^{n-1}] (1+z)^{2n-1}
\frac{1}{(1-(1+z)(1-(1-z)/(1+z))/2)^2}
\frac{1+z}{1-z}
\\ = [z^{n-1}] (1+z)^{2n-1}
\frac{1}{(1-(1+z-(1-z))/2)^2}
\frac{1+z}{1-z}
\\ = [z^{n-1}] (1+z)^{2n}
\frac{1}{(1-z)^3}.$$
Second part
We get for the argument to the absolute value being negative the
contribution
$$- \sum_{p=0}^{n-1} \sum_{q=n-p}^n
(n-p-q) {n+p-q\choose p} {n-p+q-1\choose q}
\\ = - \sum_{p=0}^{n-1} \sum_{q=0}^p
(n-p-(n-q)) {n+p-(n-q)\choose p} {n-p+n-q-1\choose n-q}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^p
(p-q) {p+q\choose p} {2n-p-q-1\choose n-q}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^p
q {2p-q\choose p} {2n-2p+q-1\choose n-p+q}
\\ = \sum_{p=0}^{n-1} \sum_{q=0}^p
q {2p-q\choose p-q} {2n-2p+q-1\choose n-1-p}.$$
We have the first part with $q$ replaced by $q+1$ and
adjust the infinite series accordingly, getting:
$$[z^{n-1}] (1+z)^{2n-1}
\sum_{p=0}^{n-1} z^p (1+z)^{-2p} [w^p] (1+w)^{2p}
\frac{(1+z)w/(1+w)}{(1-(1+z)w/(1+w))^2}
\\ = [z^{n-1}] (1+z)^{2n}
\sum_{p\ge 0} z^p (1+z)^{-2p} [w^{p-1}] (1+w)^{2p+1}
\frac{1}{(1-wz)^2}.$$
We extended $p$ to infinity due to the extractor in $z.$
Unfortunately this isn't quite a repeat so we need to do
the residues one more time. We have
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{p}} (1+w)^{2p+1} \frac{1}{(1-wz)^2}.$$
Now again put $w/(1+w)=v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$
to get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{p}} \frac{1}{(1-v)^{p+1}}
\frac{1}{(1-vz/(1-v))^2} \frac{1}{(1-v)^2}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{p}} \frac{1}{(1-v)^{p}}
\frac{1}{(1-v-vz)^2} \frac{1}{1-v}.$$
Next put $v = (1-\sqrt{1-4u})/2$ so that $v(1-v)=u$ and $dv =
1/\sqrt{1-4u} \; du$ to obtain
$$\;\underset{u}{\mathrm{res}}\;
\frac{1}{u^{p}}
\frac{1}{(1-(1+z)(1-\sqrt{1-4u})/2)^2}
\frac{1}{\sqrt{1-4u}}
\frac{1-\sqrt{1-4u}}{2u}.$$
Evaluate at $u=z/(1+z)^2$ to cancel the remaining sum (substitution
rule, same as before),
$$[z^{n-1}] (1+z)^{2n}
\frac{1}{(1-(1+z)(1-(1-z)/(1+z))/2)^2}
\frac{1+z}{1-z} \frac{1-(1-z)/(1+z)}{2}
\\ = [z^{n-1}] (1+z)^{2n}
\frac{1}{(1-(1+z-(1-z))/2)^2}
\frac{z}{1-z}
\\ = [z^{n-1}] (1+z)^{2n}
\frac{z}{(1-z)^3}
= [z^{n-2}] (1+z)^{2n}
\frac{1}{(1-z)^3}.$$
Conclusion
It remains to add up the two pieces. We have
$$[z^{n-1}] (1+z)^{2n}
\frac{1}{(1-z)^3}
+ [z^{n-1}] (1+z)^{2n}
\frac{z}{(1-z)^3}
\\ = [z^{n-1}] (1+z)^{2n+1} \frac{1}{(1-z)^3}.$$
This is
$$S = \;\underset{z}{\mathrm{res}}\;
\frac{1}{z^n} (1+z)^{2n+1} \frac{1}{(1-z)^3}.$$
Residues sum to zero so we may evaluate using minus the residues at
$z=1$ and at infinity. We require for the former (flip sign one more
time)
$$\frac{1}{2}
\left.\left[ \frac{1}{z^n} (1+z)^{2n+1} \right]''\right|_{z=1}
\\ = \frac{1}{2}
\left.\left[ - \frac{n}{z^{n+1}} (1+z)^{2n+1}
+ \frac{1}{z^{n}} (2n+1) (1+z)^{2n} \right]'\right|_{z=1}
\\ = \frac{1}{2}
\left[ n(n+1) 2^{2n+1} - n(2n+1) 2^{2n}
- n (2n+1) 2^{2n} + (2n+1)(2n) 2^{2n-1} \right]
\\ = \frac{1}{2} n 4^n.$$
We get from the residue at infinity with the sign flipped
$$\;\underset{z}{\mathrm{res}}\; \frac{1}{z^2}
z^n (1+1/z)^{2n+1} \frac{1}{(1-1/z)^3}
\\ = \;\underset{z}{\mathrm{res}}\; \frac{1}{z^2}
z^n (1+z)^{2n+1} \frac{1}{z^{2n+1}} \frac{z^3}{(z-1)^3}
= - \;\underset{z}{\mathrm{res}}\; \frac{1}{z^n}
(1+z)^{2n+1} \frac{1}{(1-z)^3} = - S.$$
We have shown that $S= \frac{1}{2} n 4^n - S$ or alternatively
$$S= n 4^{n-1}.$$
Asymptotics
OP asks to scale $S$ by the central binomial
coefficient. This will produce
$${2n\choose n}^{-1} n 4^{n-1}
\sim \frac{\sqrt{\pi n}}{4^n} n 4^{n-1}
= \frac{1}{4} \sqrt{\pi} n^{3/2}.$$
Acknowledgement
I would like to dedicate this computation to G. Egorychev
who sadly is no longer with us as of December 2023.
