Well, make it an answer. The simple observation, also true for highly composite numbers (which resemble this problem) is that getting a large value of $p(n)$ means $n$ has non-increasing  exponents in its prime factorization. Furthermore, no exponent is exactly $1.$ 

So, I am not entirely convinced that $p(m(5)) = 1296.$ Maybe, maybe not. However, i am convinced that if you go up to eight primes,
$$ n = 2^a 3^b 5^c 7^d 11^e 13^f 17^g 19^h $$
with $a \geq b \geq c \geq d \geq e \geq f \geq g \geq h \geq 0$ and none exactly 1, say $a \leq 6,$ say, and loops with built-in bounds reflecting $n > 2 \cdot 10^8,$ you will find the winner.

Oh, if $h=0,$ for example, you do not multiply that in. The HC and superabundant numbers behave better in this regard. Put another way, your $p$ is almost multiplicative, but not unless you take care to build in $p(1) = 1.$


EDITTTTT: so far, the smallest i have is     
$$  2^6 3^6 5^4 7^3 11^3 \approx e^{30.2197} \approx 1.3 \cdot 10^{13}$$

BETTER   
$$  2^9 3^6 5^4 7^3 11^2 \approx e^{29.9013} \approx 9.7 \cdot 10^{12}$$