The following gives a very weak bound. Set $h(n) = \frac{\sigma(n)}{n}$, and $$H(n) = \prod_{p|n} \frac{p}{p-1}.$$ Note that it is not hard to show that $h(n) \leq H(n)$ with equality if and only if $n=1$. This inequality is best possible in the sense that $\lim_{k \rightarrow \infty } h(n^k) = H(n)$, so $H(n)$ is the best bound we can have when we only know the prime factors of $n$. \

There is a constant $c$ such that for all sufficiently large colossally abundant $n$, we have $\sigma(n) \geq c n \log \log n$. So
$$H(n) \geq h(n) = \frac{\sigma(n)}{n} = c \log \log n. $$
Now, we know that $$H(n) = \prod_{p \leq x} \frac{p}{p-1}$$ for some $x$. We can use Mertens' Theorem to estimate that $e^\gamma \log x \approx \log \log n$. So our minimum $x$ is about $(\log n)^{\frac{1}{e^\gamma}}$. By the prime number theorem then
$$\prod_{p|n} p \approx e^{(\log n)^{\frac{1}{e^\gamma}}}.$$

This will give you an upper bound on $a_1$ by giving you a weak lower bound on the product of all the distinct prime divisors of $n$. I haven't gone through and been careful about the exact error bounds, but since there are a lot of explicit bounds on the PNT and Merten's Theorem such as those of Dusart, and earlier those of Rosser and Schoenfeld, one should be able to get an explicit bound. Since this doesn't deal with the fact that for any fixed $i$, for large $n$, $a_2, a_3 \cdots a_i$ will themselves all be large, this is likely a pretty severe underestimate.