The condition $2mnp = (m + 2)(n + 2)(p + 2)$ rewrites as $$\begin{aligned} &\left(1 + \frac{2}{m}\right)\left(1 + \frac{2}{n}\right)\left(1 + \frac{2}{p}\right) \\ &= 2. \end{aligned}$$ If $m = 1$ the first factor alone is $3 \gt 2,$ and if $m = 2$ it equals $2$ while the other factors exceed $1;$ both are impossible. If $m \ge 5,$ then since $n \ge m$ the first two factors are at most $\left(\frac{7}{5}\right)^2 = \frac{49}{25},$ forcing $1 + \frac{2}{p} \ge \frac{50}{49},$ i.e. $p \le 98.$

For $m = 4$ the equation becomes $4np = 3(n + 2)(p + 2),$ i.e. $(n - 6)(p - 6) = 48,$ so $p \le 54.$ For $m = 3$ it becomes $6np = 5(n + 2)(p + 2),$ i.e. $np - 10n - 10p - 20 = 0,$ or $(n - 10)(p - 10) = 120.$ Both factors must be positive (if $n, p \lt 10$ the product $(10 - n)(10 - p)$ is at most $49$), so the largest $p$ comes from $n - 10 = 1:$ $n = 11$ and $p = 130.$ Indeed $2 \cdot 3 \cdot 11 \cdot 130 = 8580$ $= 5 \cdot 13 \cdot 132.$

Since every other case yields $p \le 98,$ the largest possible value is $p = 130.$