Let $(W,S)$ be a Coxeter system. Let $R(w)$ be the number of reduced words for an element, let $D(w)\subseteq S$ be the set of right descents of $w$. Then
$$R(w)=\sum_{s\in D(w)}R(ws)$$
with $R(1)=1$.

This recurrence relation is not how you want to compute the number of reduced words for the longest element of $S_n$, though, because there is a closed formula for this, and using the recurrence relation would take forever.

If you want to build $R(w_0(n))$ from $R(w_0(n-1))$, note that the closed formula for these numbers comes from the hook length formula for standard tableaux on the shape $(n,n-1,\ldots,2,1)$. The product of the hook lengths of every box except for those in the first row is
$$\frac{\binom{n}2!}{R(w_0(n-1))}$$
The remaining hook lengths are $2n-1$, $2n-3$, $\ldots$, $3$, $1$. So
$$\frac{\binom{n+1}2!}{R(w_0(n))}=(2n-1)!!\frac{\binom n2 !}{R(w_0(n-1))}$$