jmc

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 $$\begin{array}{rl}\frac{-(-8)\pm \sqrt{(-8{)}^2-4(1)(5-k)}}{2(1)}& =\frac{8\pm \sqrt{64-20+4k}}{2}\\ & =4\pm \sqrt{11+k}.\end{array}$$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 $$\begin{array}{rl}\sqrt{3(11+k)}& =k+11\Rightarrow \\ 3(11+k)& =(k+11{)}^2\Rightarrow \\ 33+3k& =k^2+22k+121\Rightarrow \\ 0& =k^2+19k+88\Rightarrow \\ 0& =(k+8)(k+11).\end{array}$$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 $k=\boxed{-8}$.