jmc

Label points $O,N,M$ as follows. Because folding preserves the corner $A$ (which is now $C$), we have two congruent triangles $△BAN\cong △BCN$. This means that $AB=BC=5$. Knowing this, we can compute $MB=8-5=3$ and $MC=\sqrt{5^2-3^2}=4$.

Notice also that we have similar triangles $△BMC\sim △CON$. (This can be determined by angle chasing.) Thus, we have $\frac{MB}{CO}=\frac{BC}{CN}=\frac{MC}{ON}=\frac{4}{8}$. Knowing $MB=3$ and $CB=5$, we can compute $CO=6$ and $CN=10$.

Now, we see that $AN=6+4=10$. By Pythagoras on $△BAN$, we have $BN=\sqrt{5^2+10^2}=5\sqrt{5}$. Hence, $l=\boxed{5\sqrt{5}}$.