Hi "DC". I think that I have worked out that the Hermite normal form is a "trichotomous quasipolynomial" in the variable $p$. If $f:\mathbb{Z} \to \mathbb{Z}$ is a function, then my definition is that $f$ is a trichotomous quasipolynomial if it is a quasipolynomial for $x \gg 0$, possibly a different quasipolynomial for $x \ll 0$, and in between finitely many unrestricted values.
I think that if $R$ is a canonically Euclidean ring — Euclidean with canonically chosen quotients and remainders — then there is a Hermite normal form for matrices over $R$. In particular, I think that $R$ does not have to be a Euclidean domain.
As a first try, let $A$ be the ring of all functions $f:\mathbb{Z} \to \mathbb{Z}$, using pointwise quotients and remainders. Hermite normal form over this ring is a model of computing Hermite normal form for any $\mathbb{Z}$ family of integer matrices. $A$ is sort-of a Euclidean ring, except it isn't Noetherian.
Let $B$ be the subring of $A$ consisting of trichotomous quasipolynomials. Then I believe that $B$ is Noetherian and it is a Euclidean ring. If that is correct, then you obtain a Hermite normal form that is also a trichotomous quasipolynomial.