Can this product be written so that symmetry is manifest?

Let $i,$ $j,$ $k$ be nonnegative integers such that $i+j+k$ is even. The expression $$(-1)^{j+k}\binom{i+j+k}{i,j,k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}$$ apparently computes the coefficients of certain terms in the expansion of a certain polynomial. (My answer to this question provides motivation. Incidentally, that question has an imminently-expiring bounty and deserves more attention - certainly something better than my own poor contribution.) It can be shown - see below - that this expression is invariant under permutation of $i,$ $j,$ $k.$ My question is, is there a natural way of rewriting the expression so that this symmetry is more apparent? Since the multinomial coefficient is manifestly symmetric, the main focus is on the remaining factors, $$g(i,j,k):=(-1)^{j+k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}.$$ If $g(i,j,k)$ equals, or is closely related to some named object, that would be very interesting to me as well.

The restriction that $i+j+k$ be even is necessary: in the case $i+j+k$ odd, $g(i,j,k)$ is, in fact, indeterminate for $k>i,j$; the sign is also changed by certain permutations in that case.

Some thoughts: We can express $g(i,j,k)$ in terms of generalized binomial coefficients as $$g(i,j,k)=(-1)^{j+k}\frac{\binom{\frac{1}{2}(i-j+k-1)}{k}}{\binom{\frac{1}{2}(i+j+k-1)}{k}}=(-1)^j\frac{\binom{\frac{1}{2}(-i+j+k-1)}{k}}{\binom{\frac{1}{2}(i+j+k-1)}{k}},$$ but this does not seem helpful.

To prove the symmetry it suffices to show that $g(i,j,k)=g(j,i,k)=g(k,j,i)$ since these two permutations generate the full permutation group. The first equality is proved as follows: $$\begin{aligned}g(j,i,k)&=(-1)^{i+k}\prod_{\ell=0}^{k-1} \frac{-i+j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=(-1)^{i+k}\prod_{\ell=0}^{k-1} \frac{-i+j-k+2\ell+1}{i+j+k-2\ell-1}\\ &=(-1)^i\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\&=g(i,j,k).\end{aligned}$$ The second line is obtained from the first by substituting $k-1-\ell$ for $\ell$ in the numerator; the last line follows since $(-1)^i=(-1)^{j+k}$ by the assumption $i+j+k$ even.

Equality of $g(i,j,k)$ and $g(k,j,i)$ is proved by taking $k\ge i$. We have $$\begin{aligned}1&=\prod_{\ell=0}^{k-i-1}\frac{i-j-k+2\ell+1}{i-j-k+2\ell+1}\\ &=\prod_{\ell=0}^{k-i-1}\frac{-i-j+k-2\ell-1}{i-j-k+2\ell+1}\\ &=\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{-i-j-k+2\ell+1}\\ &=(-1)^{k-i}\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}.\end{aligned}$$ The second line is obtained by substituting $k-i-1-\ell$ for $\ell$ in the numerator; the third is obtained by substituting $\ell-i$ for $\ell$ in both numerator and denominator. Then $$\begin{aligned}g(k,j,i)&=(-1)^{i+j}\prod_{\ell=0}^{i-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=(-1)^{i+j}\prod_{\ell=0}^{i-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}(-1)^{k-i}\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=g(i,j,k).\\ \end{aligned}$$

Let $\mathbb O$ be the set of all odd integers. Define a function $Q:\mathbb O\to\mathbb Q$ by requiring

$Q\left(2n-1\right) = \left\lbrace \begin{array}{c} \prod\limits_{i=1}^{n}\left(2i-1\right),\text{ if }n > 0; \\ \prod\limits_{i=n+1}^{0}\dfrac{1}{2i-1},\text{ if }n \leq 0 \end{array} \right.$ for every $n\in\mathbb Z$.

Thus, every $m\in\mathbb O$ satisfies $Q\left(m\right) = m Q\left(m-2\right)$, and we have $Q\left(-1\right)=Q\left(1\right)=1$.

(Note how, unlike the factorial function, $Q$ is well-defined both on the positive and the negative side.)

Then,

$g\left(i,j,k\right) = \left(-1\right)^{\left(i+j+k\right)/2}\dfrac{Q\left(i+j-k-1\right)Q\left(j+k-i-1\right)Q\left(k+i-j-1\right)}{Q\left(i+j+k-1\right)}$.