Mr Helms,
This is the $n=1$ case. Your formula gives $e_{1,q}=q$. Say we want to study how often prime $q=q_k$ divides $\prod_{p \leq x}(p-1)$. Maybe write this product as $$ \left(\prod_{i=1}^m\prod_{\substack{p \leq x\\ p \in (q^{i-1}\mathbb{Z}+1)\setminus(q^{i}\mathbb{Z}+1)}}(p-1)\right) \times \prod_{\substack{p \leq x\\ p \in (q^{m}\mathbb{Z}+1)}}(p-1). $$ If $m$ is the right size relative to $x$, then counting primes $p$ in $(q^{i-1}\mathbb{Z}+1)\setminus(q^{i}\mathbb{Z}+1)$, $i \leq m$ can be done by an asymptotic version of Dirichlet's theorem on primes in arithmetic progressions. (Siegel-Walfisz theorem)
Equating $\prod_{p \leq x}(p-1)$ and $\prod_{k=1}^{w}(p_k-1)$, we get $w \approx x/\log x$.
If $q^i \ll (\log x)^N$, as required by Siegel-Walfisz theorem, then the number of primes $p \leq x$ in $(q^{i-1}\mathbb{Z}+1)\setminus(q^{i}\mathbb{Z}+1)$ is $\frac{q-1}{\varphi(q^i)}\frac{x}{\log x} + O\left(x \exp(-c_N (\log x)^{1/2})\right) = \frac{x}{q^{i-1}\log x} + O\left(x \exp(-c_N (\log x)^{1/2})\right)$. The number of primes $p \leq x$ in $q^m\mathbb{Z}+1$ is $\frac{x}{q^{m-1}(q-1)\log x}+O\left(x \exp(-c_N (\log x)^{1/2})\right)$. So, a lower bound for $a_{1,k}/w$ is
$$ \left(\left(\frac{m}{q^{m-1}(q-1)}+\sum_{i=1}^m\frac{i-1}{q^{i-1}}\right)\frac{x}{\log x} + O\left(x \exp(-c_N (\log x)^{1/2})\right)\right)/\left(x/\log x\right) $$ where $q^m \ll (\log x)^N$. If we approximate
$$ \frac{m}{q^{m-1}(q-1)}+\sum_{i=1}^m\frac{i-1}{q^{i-1}} $$
with
$$ \sum_{i=1}^{\infty}\frac{i-1}{q^{i-1}} = \frac{q}{(q-1)^2} $$
and take $x$ to $\infty$, then this approximation for $a_{1,k}/w$ agrees with your formula for $e_{1,q}$.