There are two solutions: x=-0.8058 and -3.1942 approximately. If we add these together we get -4. But see the easier solution below.
First, let y=x^2+4x+5=x^2+4x+4+1=(x+2)^2+1. Now go down to EASIER SOLUTION.
So y^(y^y)=2017
When x=-2, y=1 and 1^(1^1)=1.
When x=-1 or -3, y=2 and 2^(2^2)=16.
When x=0 or -4, y=5 and 5^(5^5)>2017, so we know that -1<x<0 or -4<x<-3, so 2<y<5.
EASIER SOLUTION
Let's suppose that y=a is the right answer to y^(y^y)=2017.
(x+2)^2=a-1, so x+2=±√(a-1) and x=-2+√(a-1) or -2-√(a-1). If we add these roots together we get -4.
(The actual solution is y=2.426049=a approx, therefore (x+2)^2=a-1=1.426049; x+2=±√1.426049, so the sum of the zeroes is 2*(-2)=-4.)