Multiplying the given equations together, we have \[\begin{aligned} (xyz)^2& = (-80-320i) \cdot 60 \cdot (-96+24i) \\ &= -80(1+4i) \cdot 60 \cdot -24(4-i) \\ &= (80 \cdot 60 \cdot 24) (8 + 15i).\end{aligned}\]To solve for $ xyz, $ we find a complex number $ a+bi $ whose square is $ 8+15i $ (where $ a $ and $ b $ are real); that is, we want \[(a+bi)^2 = (a^2-b^2) + 2abi = 8 + 15i.\]Equating the real and imaginary parts, we get the equations $ a^2-b^2=8 $ and $ 2ab=15 $. Then $ b = \frac{15}{2a}; $ substituting into the other equation gives $ a^2 - \frac{225}{4a^2} = 8, $ or $ 4a^4 - 32a^2 - 225 = 0 $. This factors as \[(2a^2-25)(2a^2+9) = 0\]so $ 2a^2-25=0 $ (since $ a $ is real), and $ a = \pm \frac{5}{\sqrt2} $. Then $ b = \frac{15}{2a} = \pm \frac{3}{\sqrt2} $. Therefore, (arbitrarily) choosing both $ a $ and $ b $ positive, we have \[(xyz)^2 = (80 \cdot 60 \cdot 24) \left(\frac{5}{\sqrt2} + \frac{3}{\sqrt2}i \right)^2,\]and so \[\begin{aligned} xyz& = \pm \sqrt{80 \cdot 60 \cdot 24}\left(\frac{5}{\sqrt2}+ \frac{3}{\sqrt2}i \right) \\&= \pm240\sqrt{2} \left(\frac{5}{\sqrt2} + \frac{3}{\sqrt2}i \right) \\ &= \pm240(5+3i).\end{aligned}\]Then \[\begin{aligned} x &= \frac{xyz}{yz} =\pm \frac{ 240(5+3i)}{60} = \pm (20 + 12i), \\ z &= \frac{xyz}{xy} =\pm \frac{ 240(5+3i)}{-80(1+4i)} = {\pm} \frac{-3 (5+3i)(1-4i)}{17} =\pm \frac{ -3(17-17i)}{17} = \pm( -3+3i), \\ y &= \frac{xyz}{xz} = {\pm}\frac{ 240(5+3i)}{-24(4-i)} = \pm \frac{- 10 (5+3i)(4+i)}{17} = \pm \frac{ -10(17+17i)}{17} = \pm(-10-10i).\end{aligned}\]Therefore, $ x+y+z = \pm(7 +5i), $ so $ |x+y+z| = \sqrt{7^2+5^2} = \boxed{\sqrt{74}} $.