A curious congruence modulo primes

For integers $l>0$ and $m\geqslant0$, we introduce the numbers $$S_l^{(m)}(n)=\sum_{k_1,\ldots,k_l\in\mathbb N\atop k_1+\cdots+k_l=n}\binom n{k_1,\ldots,k_l}^m \ \ (n=0,1,2,\ldots),$$ and prove that for any prime $p$ not dividing $l+1$ we have the congruence $$\sum_{n=1}^{p-1}\frac{(-1)^{mn}}{n^{m-1}}S_l^{(m)}(n)\equiv0\pmod p.$$ When $l=4$ and $m=2$, this yields the curious congruence $$\sum_{n=1}^{p-1}\frac{D(n)}n\equiv0\pmod p$$ for any prime $p\not=5$, where the Domb number $D(n)$ is given by $$D(n)=\sum_{k=0}^n\binom nk^2\binom{2k}k\binom{2(n-k)}{n-k}.$$