jmc

First consider the points in the six parallelepipeds projecting 1 unit outward from the original parallelepiped. Two of these six parallelepipeds are 1 by 3 by 4, two are 1 by 3 by 5, and two are 1 by 4 by 5. The sum of their volumes is $2(1\cdot 3\cdot 4+1\cdot 3\cdot 5+1\cdot 4\cdot 5)=94$. Next consider the points in the twelve quarter-cylinders of radius 1 whose heights are the edges of the original parallelepiped. The sum of their volumes is $4\cdot \frac{1}{4}\pi \cdot 1^2(3+4+5)=12\pi$. Finally, consider the points in the eight octants of a sphere of radius 1 at the eight vertices of the original parallelepiped. The sum of their volumes is $8\cdot \frac{1}{8}\cdot \frac{4}{3}\pi \cdot 1^3=\frac{4\pi }{3}$. Because the volume of the original parallelepiped is $3\cdot 4\cdot 5=60$, the requested volume is $60+94+12\pi +4\pi /3=\frac{462+40\pi }{3}$, so $m+n+p=462+40+3=\boxed{505}$.