One vertex of the triangle is on the vertex of the parabola. The $ x $-coordinate of the vertex is $ \frac{-b}{2a}=\frac{-(-8)}{2(1)}=4 $. To find the $ y $-coordinate, we plug in $ x=4 $ to find $ y=4^2-8\cdot 4+5=16-32+5=-11 $. So one vertex of the triangle is at $ (4, -11) $. The other two vertices are on the intersection of the parabola $ y=x^2-8x+5 $ and the line $ y=k $. Thus we have $ x^2-8x+5=k $ or $ x^2-8x+(5-k)=0 $. By the quadratic formula, the solutions to this equation are \begin{align*} \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(5-k)}}{2(1)}&=\frac{8\pm\sqrt{64-20+4k}}{2}\\ &=4\pm\sqrt{11+k}.\end{align*}So the two other vertices of the triangle are $ (4-\sqrt{11+k},k) $ and $ (4+\sqrt{11+k},k) $. Now, we know the triangle is equilateral. Since two vertices are on the same horizontal line, the side length is the difference of their $ x $-coordinates, which is $ (4+\sqrt{11+k})-(4-\sqrt{11+k})=2\sqrt{11+k} $. The height of the equilateral triangle is $ \frac{\sqrt{3}}{2} $ times the side length, which is $ \frac{\sqrt{3}}{2}(2\sqrt{11+k})=\sqrt{3(11+k)} $. But the height is also the difference in the $ y $-coordinate between the vertex and the horizontal side which is at $ y=k $. This means the height is equal to $ k-(-11)=k+11 $, since $ -11 $ is the $ y $-coordinate of the vertex. These heights must be equal, so we can write the equation \begin{align*} \sqrt{3(11+k)}&=k+11\quad\Rightarrow\\ 3(11+k)&=(k+11)^2\quad\Rightarrow\\ 33+3k&=k^2+22k+121\quad\Rightarrow\\ 0&=k^2+19k+88\quad\Rightarrow\\ 0&=(k+8)(k+11).\end{align*}Thus we have $ k=-8 $ or $ k=-11 $. We can throw out $ k=-11 $ because then the line $ y=-11 $ intersects the parabola only once, at the vertex, so there's no triangle, just a point. Thus we have $ k=\boxed{-8} $.