There is a different formula for $\#\mathrm{Aut}(\prod_i C_{p^{n_i}}^{k_i})$ given in the paper Automorphisms of Finite Abelian Groups by Hillar and Rhea:
$$\#\mathrm{Aut}(\prod_{t=1}^m C_{p^{e_t}}) = \prod_{t=1}^m (p^{d_t} - p^{t-1}) p^{e_t(m-d_t) + (e_t-1)(m-c_t+1)},$$
where $1\leq e_1\leq e_2\leq \cdots\leq e_m$, and $c_t$ and $d_t$ are the minimum and maximum of the set $S_t := \{\ell\ :\ e_\ell=e_t\}$, respectively.
Below I will show that OP's formula is equivalent to the Hillar-Rhea formula.
Let $d_0:=0$. It can be seen that the $k_i$'s are the nonzero elements of the multiset $\{ d_1-d_0, d_2-d_1, \dots, d_m-d_{m-1}\}$ and the $n_i$'s are the corresponding elements of $\{e_1,e_2,\dots,e_m\}$. Define $s_0=0, s_1, \dots, s_q$ be the indices such that $k_i = d_{s_i} - d_{s_{i-1}}$ and $n_i = e_{s_i}$.
Vice versa, $d_{s_i} = k_1+\dots+k_i$ and $c_{s_i} = d_{s_{i-1}}+1$.
First consider these parts of the two formulas:
$$\prod_{i=1}^q \prod_{j=0}^{k_i-1} (p^{k_i} - p^j) = \prod_{i=1}^q p^{k_i(k_i-1)/2} \prod_{j=0}^{k_i-1} (p^{k_i-j} - 1)$$
and
$$\prod_{t=1}^m (p^{d_t} - p^{t-1}) = p^{m(m-1)/2}\prod_{t=1}^m (p^{d_t-t+1} - 1).$$
It is easy to see that the multisets $\{ k_i - j : 0\leq j \leq k_i-1, 1\leq i\leq q \}$ and $\{ d_t - t +1\ :\ 1\leq t\leq m \}$ are the same, since the $t$-th element in the sequence
$$k_1 - 0, k_1 - 1, \dots, 1, k_2 - 0, k_2 - 1, \dots, 1, \dots$$
equals $d_t-t+1$.
Now it remains to prove the equality for the powers of $p$ in the two formulas, i.e.
$$\sum_{i=1}^q \bigg(k_i(k_i-1)/2 + (n_i-1)k_i^2 + \sum_{j\ne i} \min(n_i,n_j)k_ik_j\bigg) = m(m-1)/2 + \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big).$$
In the l.h.s. we have
$$\sum_{i=1}^q \sum_{j\ne i} \min(n_i,n_j)k_ik_j = 2\sum_{i=1}^q n_i k_i \sum_{j>i} k_j=2\sum_{i=1}^q n_i k_i (m-d_{s_i}).$$
In the r.h.s. we have
\begin{split}
\sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big) &= \sum_{i=1}^q k_i\big(e_{s_i}(m-d_{s_i}) + (e_{s_i}-1)(m-d_{s_{i-1}})\big) \\
&= \sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m-d_{s_{i-1}})\big)\\
&=\sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m+k_i-d_{s_i})\big)\\
&=2\sum_{i=1}^q k_i n_i(m-d_{s_i}) + \sum_{i=1}^q \big( (n_i-1)k_i^2 - (m-d_{s_i})k_i\big).
\end{split}
Finally, we notice that
$$\sum_{i=1}^q k_i(k_i-1)/2 = m(m-1)/2 - \sum_{i=1}^q (m-d_{s_i})k_i$$
since $m=k_1+k_2+\dots+k_q$ and $m-d_{s_i} = k_{i+1}+k_{i+1}+\dots+k_q$. QED
So, we can conclude that OEIS A034382 did indeed contain an error in its 16-th term. Now it's corrected.
Best Answer
Your proposed formula is indeed correct. The numbers of finite abelian groups of fixed order are given by OEIS A000688.