Let $\mathrm{gnu}(n)$ denote the number of groups of order $n$, and $p$ denote an odd prime.
It's well known that $$\mathrm{gnu}(2p)=2,$$
and that
$$\mathrm{gnu}(4p)= \begin{cases}5 & p \equiv 1 \mod 4 \\4 & p \equiv 3 \mod 4\end{cases}$$
with the one exception that $\mathrm{gnu}(4 \cdot 3) = 5$.
Looking through the data of the number of groups of order $8p$, $16p$, it looks to me that
$$\mathrm{gnu}(8p)= \begin{cases}15 & p \equiv 1 \mod 8 \\12 & p \equiv 3,7 \mod 8\\14 & p \equiv 5 \mod 8\end{cases}$$
with the exceptions $\mathrm{gnu}(8 \cdot 3) = 15$, $\mathrm{gnu}(8 \cdot 7) = 13$, and
$$\mathrm{gnu}(16p)= \begin{cases}54 & p \equiv 1 \mod 16 \\42 & p \equiv 3,7,11,15 \mod 16\\51 & p \equiv 5,13 \mod 16\\53 & p \equiv 9 \mod 16\end{cases}$$
with the exceptions $\mathrm{gnu}(16 \cdot 3) = 52$, $\mathrm{gnu}(16 \cdot 5) = 52$, $\mathrm{gnu}(16 \cdot 7) = 43$, although I haven't attempted to prove either of these rigorously.
Altogether, it looks to me like $\mathrm{gnu}(2^n \cdot p)$ depends on $p$ modulo $2^n$, and takes on a constant value, other than a few small exceptions. Is this known to be true? If not, is a counterexample known? (Also, are my results for $\mathrm{gnu}(8 p)$ and $\mathrm{gnu}(16 p)$ actually correct?)
Edit: If you're curious, I've just written a computer program to check through data on groups of order up to $50,000$ (see this) to attempt to find patterns for $32p$ and $64p$, and indeed it did. Here's a link to a pastebin, I won't write it all out here, to save making the question even longer. It couldn't really do $128p$ with confidence, since there's only one group of the relevant orders in each of the classes at that point.
Let $G$ be a group of order $2^np$. If $p$ is large enough with respect to $2^n$, then a Sylow $p$-subgroup of $G$ will be normal, so $G$ will be a semi-direct product $C_p\rtimes G_2$, with $G_2$ a Sylow $2$-subgroup. So the number of isomorphism types will depend only on the number of choices for $G_2$ and the number of choices of homomorphisms $G_2\to \mathrm{Aut}(C_p)$ (up to equivalence). It is well-known that $\mathrm{Aut}(C_p)\cong C_{p-1}$ which has a normal $2$-subgroup $N$ which must then contain $\mathrm{im}(G_2)$. Moreover $N$ is cyclic so it has a unique subgroup of each order, so $\mathrm{im}(G_2)$ is contained in the unique subgroup of order $|G_2|$ of $N$. So the number of choices only depends on the highest power of $2$ dividing $p-1$, up to $|G_2|$.
(This also explains your data: the answer is bigger as the power of $2$ dividing $p-1$ increases. For example you get a maximum when $p\equiv 1 \pmod {2^n}$.)