jmc

First we distribute the constants in $4(x^2-2x+2)-7(x^3-3x+1)$ to get $4x^2-8x+8-7x^3+21x-7.$ Combining like terms, we find that this is $-7x^3+4x^2+13x+1.$ Then, the sum of the squares of all the coefficients is $(-7{)}^2+(4{)}^2+(13{)}^2+(1{)}^2=49+16+169+1=\boxed{235}.$