Ok, I think I've found an answer along with a proof. I will go through a series of claims (of course there may be more efficient ways of going about this). The main result is in Claim 3.
Claim 1. For nonnegative integers $m, r, t$ we have
$$ \sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} = (-1)^{m+t}\binom{r}{m}. \tag*{$(1)$} $$
Proof. We proceed by Egorychev's method. We will use the fact that
$$ \binom{j+r}{t+r} = \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz, $$
where the integral is a contour integral over a circle $|z|=\varepsilon$ for some $0<\varepsilon<\infty$.
Using this fact, and interchanging the summation and integral signs when necessary, we have
\begin{align*}
\sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} &= \sum_{j=0}^{m+t} \binom{m+t}{j} \left( \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz \right) (-1)^{j} \\[1.2ex]
&= \frac{1}{2\pi i}\oint \; \sum_{j=0}^{m+t} \binom{m+t}{j} \frac{(1+z)^{j+r}}{z^{t+r+1}} (-1)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (1+z)^{j} (-1)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (-1-z)^{j} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (1 - 1 - z)^{m+t} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (-z)^{m+t} \, dz \\[1.2ex]
&= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{r-m+1}} (-1)^{m+t} \, dz \\[1.2ex]
&= (-1)^{m+t}\binom{r}{r-m} \\[1.2ex]
&= (-1)^{m+t}\binom{r}{m}.
\end{align*}
$$\tag*{$\blacksquare$}$$
Claim 2. For nonnegative integers $n, r, s$ with $s\ge r$, we have
$$ \sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{r}{n-s}. \tag*{$(2)$} $$
Proof. If $s > n$, then both sides of $(2)$ are clearly $0$, so for the proof let us assume $s\le n$. Take $m = n-s$ and $t = s-r$ and plug these into $(1)$. We obtain
$$ \sum_{j=0}^{n-r} \binom{n-r}{j}\binom{j+r}{s}(-1)^{j} = (-1)^{n-r}\binom{r}{n-s}. $$
We can shift the summation index by taking $j = k-r$ (where $k$ is the new index). By doing this and then multiplying both sides by $(-1)^{r}$, we get $(2)$.
$$\tag*{$\blacksquare$}$$
Claim 3. For nonnegative integers $n, r, s$, we have
$$ \boxed{ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} } $$
where the RHS involves the multinomial coefficient.
Proof. Without loss of generality, assume $s\ge r$. Consider the fact that
$$ \binom{n}{k}\binom{k}{r} = \binom{n}{r}\binom{n-r}{k-r} $$
and use this to obtain
$$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = \binom{n}{r}\sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k}. $$
By Claim 2, we see that the RHS reduces to
$$ \binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s}. $$
This is then decomposed as
\begin{align*}
\binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s} &= (-1)^{n} \frac{n!}{r!(n-r)!}\frac{r!}{(n-s)!(r-n+s)!} \\
&= (-1)^{n}\frac{n!}{(n-r)!(n-s)!(r+s-n)!},
\end{align*}
and the fraction in the last expression can be identified as a multinomial coefficient, giving us
$$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} $$
as desired.
$$\tag*{$\blacksquare$}$$
Best Answer
We can rewrite the sum as:
$$\begin{aligned}S&=\sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose 2r-1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 {n \choose n-2r+1} \\&= \sum_{r \in \mathbb{N}} (n-2r+1)^2 \frac{n!}{(n-2r+1)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) \frac{(n-1)!}{(n-2r)!(2r-1)!} \\&= \sum_{r \in \mathbb{N}} n(n-2r+1) {n-1 \choose 2r-1} \\&= n^2 \sum_{r \in \mathbb{N}} {n-1 \choose 2r-1} - n \sum_{r \in \mathbb{N}} (2r-1){n-1 \choose 2r-1} \\&= n^2 \frac{2^{n-1}}{2} - n(n-1) \frac{2^{n-2}}{2} \\&= n2^{n-3}(2n-n+1) \\&= n(n+1)2^{n-3}\end{aligned}$$