jmc

Adding the equations, we get $$ab+ac+bc+4(a+b+c)=-48.$$Multiplying the equations by $c,$ $a,$ $b,$ respectively, we get $$\begin{array}{rl}abc+4bc& =-16c,\\ abc+4ac& =-16a,\\ abc+4ab& =-16b.\end{array}$$Adding all these equations, we get $$3abc+4(ab+ac+bc)=-16(a+b+c).$$Then $$\begin{array}{rl}3abc& =-4(ab+ac+bc)-16(a+b+c)\\ & =-4(ab+ac+bc+4(a+b+c))\\ & =(-4)(-48)=192,\end{array}$$so $abc=\boxed{64}.$