The second equation is $ab + c(a + b) = -4,$ giving $ab = 3c - 4;$ group the third as $ab(c + d) + cd(a + b) = 14$

Eliminating $d$ and $cd$ leaves a quartic in $c$ whose real roots are $-2$ and $\frac{10}{3};$ one of them makes $a$ and $b$ non-real

Since $a + b = -3,$ the second equation reads $ab + c(a + b) = ab - 3c = -4,$ so $ab = 3c - 4.$ Grouping the third equation as $ab(c + d) + cd(a + b) = 14$ gives $(3c - 4)(c + d) - 3cd = 14,$ which simplifies to $3c^2 - 4c - 4d = 14,$ so $d = \frac{3c^2 - 4c - 14}{4}.$ The fourth equation becomes $(3c - 4)\,cd = 30.$

Substituting for $d$ yields $c(3c - 4)(3c^2 - 4c - 14) = 120,$ i.e. $$\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}$$ The quadratic factor has negative discriminant, so $c = -2$ or $c = \frac{10}{3}.$ If $c = \frac{10}{3},$ then $ab = 6$ with $a + b = -3,$ impossible for real $a, b$ since $9 - 24 \lt 0.$ So $c = -2,$ giving $ab = -10$ and $d = \frac{12 + 8 - 14}{4} = \frac{3}{2}.$

Then $a^2 + b^2$ $= (a + b)^2 - 2ab$ $= 9 + 20 = 29$ and $c^2 + d^2 = 4 + \frac{9}{4} = \frac{25}{4},$ so $a^2 + b^2 + c^2 + d^2 = \frac{141}{4}$ and $m + n = 141 + 4 = 145.$