The following formula $$(n,p,q,r\ \text{odd})\quad \sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\ \frac{1}{2} & \text{if exactly two indices are equal}\\ \frac{1}{3!}& \text{if}\ p=q=r= \frac{n}{3} \end{cases}$$ reminds me very much of the "multinomial formula" $$ \left( x_1 + x_2 + \cdots + x_d \right)^n = \sum_{\left|\boldsymbol{\alpha}\right|=n} \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}}\, \mathbf{x}^{\boldsymbol{\alpha}} $$ where $\boldsymbol{\alpha}\in \mathbb{N}^d,\ \left|\boldsymbol{\alpha}\right| = \alpha_1 + \alpha_2 + \cdots + \alpha_d $ and $$ \begin{split} \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}} & = \genfrac{(}{)}{0pt}{0}{n}{\alpha_1} \genfrac{(}{)}{0pt}{0}{n-\alpha_1}{\alpha_2}\, \cdots \, \genfrac{(}{)}{0pt}{0}{n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1}}{\alpha_d}= \frac{n!}{\alpha_1 !\, (n-\alpha_1)!} \frac{(n-\alpha_1)!}{\alpha_2 !\, (n-\alpha_1-\alpha_2)!}\, \cdots\, \frac{(n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1})!}{\alpha_d !\, 0 !} \\ & = \frac{n!}{\alpha_1 !\, \alpha_2 ! \, \cdots \, \alpha_d!} \end{split} $$ is the number of times the facteur $x^{\boldsymbol{\alpha}}:= x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_d^{\alpha_d}$ appears in the expansion of $\left( x_1 + x_2 + \cdots + x_d \right)^n$.
In our case $d=3,\ \boldsymbol{\alpha}= (p,q,r)$.
There could be other possibilities but I'm thinking of taking $(x_1,x_2,x_3)= \left(1, \frac{1}{2}, \frac{1}{3} \right)$ then I need to find a "simple" function of $\boldsymbol{\alpha}$, $f:\mathbb{N}^3 \to \mathbb{N}^3 $ which would take value $(1,0,0)$ if $p<q<r$ (i.e. if $\boldsymbol{\alpha}$ view as a function $1\mapsto \alpha_1=p,\ 2\mapsto \alpha_2=q,\ 3\mapsto \alpha_3=r$ is injective), take value $(0,1,0)$ if exactly two indices are equal (or if $\boldsymbol{\alpha}$ view as a the previous function takes only two values) and $(0,1,1)$ if $p=q=r$.
If we find such a function then the first sum can be rewritten $$\sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \mathbf{x}^{f(\boldsymbol{\alpha})}$$
The final step, if possible would be to relate $\mathbf{x}^{f(\boldsymbol{\alpha})}$ to $\mathbf{x}^{\boldsymbol{\alpha}}$ by integration or derivation or maybe something different.