Divide both sides of the given equality by $m_1 m_2 m_3$ and set $a = \frac{n_1}{m_1}$, $b = \frac{n_2}{m_2}$ and $c = \frac{n_3}{m_3}$. The equality $(m_1 - n_1)(m_2 - n_2)(m_3 - n_3) = m_1 m_2 m_3 - n_1 n_2 n_3$ becomes
$$(1 - a)(1 - b)(1 - c) = 1 - abc \iff a + b + c = ab + bc + ca$$
and we have to show that
$$(a + 1)(b + 1)(c + 1) \geq 8 \iff a + b + c + ab + ba + ca + abc \geq 7. \quad (1)$$
Let $a + b + c = ab + bc + ca = t$. Since $(a + b + c)^2 \geq 3(ab + bc + ca)$, we have $t^2 \geq 3t$, i.e. $t \geq 3$. Furthermore
$$a^3 + b^3 + c^3 \geq \frac{(a^2 + b^2 + c^2)^2}{a + b + c} = \frac{(t^2 - 2t)^2}{t} = t(t - 2)^2$$
and
$$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = t(t^2 - 3t) + 3abc$$
imply $3abc \geq t(t-2)^2 - t^2(t-3) = 4t - t^2$. Since (1) is equivalent to $2t + abc \geq 7$, which is true for $t \geq \frac{3}{2}$, it suffices to show that $2t + \frac{4t - t^2}{3} \geq 7$ for $t \in [3, \frac{3}{2}]$. The latter is equivalent to $(t-3)(t-7) \leq 0$, which is true for $t \in [3, 7]$.