I assume that all $h_i$ are even. Then $(n+h_1,n+h_2,\dots,n+h_k)$ is a prime tuple iff $$\begin{cases} h_1!m \equiv -1\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -1\pmod{n+h_k}, \end{cases} $$ where $m=(n-1)!$. The system implies $$\begin{cases} h_k!m \equiv -\frac{h_k!}{h_1!}\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -\frac{h_k!}{h_k!}\pmod{n+h_k}, \end{cases} $$ which further combines into $$h_k!m \equiv - \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}\pmod{(n+h_1)\cdots (n+h_k)}.$$ That is, $(n+h_1)\cdots (n+h_k)$ divides the numerator of $$h_k!(n-1)! + \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}.$$

Example. For twin primes $(n,n+2)$, we have $k=2$ with $h_1=0$ and $h_2=2$. Then the last expression becomes $$2(n-1)!+2\frac{n+2}2 + \frac{n}{-2} = \frac{4(n-1)!+n-4}{2},$$ and thus we want $n(n+2)\mid (4(n-1)!+n-4)$.

I assume that all $h_i$ are even. Then $(n+h_1,n+h_2,\dots,n+h_k)$ is a prime tuple iff $$\begin{cases} h_1!m \equiv -1\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -1\pmod{n+h_k}, \end{cases} $$ where $m=(n-1)!$. The system implies $$\begin{cases} h_k!m \equiv -\frac{h_k!}{h_1!}\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -\frac{h_k!}{h_k!}\pmod{n+h_k}, \end{cases} $$ which further combines into $$h_k!m \equiv - \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}\pmod{(n+h_1)\cdots (n+h_k)}.$$ That is, $(n+h_1)\cdots (n+h_k)$ divides the numerator of $$h_k!(n-1)! + \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}.$$

Example. For twin primes $(n,n+2)$, we have $k=2$ with $h_1=0$ and $h_2=2$. Then the last expression becomes $$2(n-1)!+2\frac{n+2}2 + \frac{n}{-2} = \frac{4(n-1)!+n+4}{2},$$ and thus we want $n(n+2)\mid (4(n-1)!+n+4)$.