jmc

By Cauchy-Schwarz, $$(a^2+b^2+c^2+d^2)(1+1+1+1)\ge (a+b+c+d{)}^2.$$Thus, $(16-e^2)(4)\ge (8-e{)}^2.$ This simplifies to $16e-5e^2\ge 0,$ or $e(16-5e)\ge 0.$ Therefore, $e\le \frac{16}{5}.$

Equality occurs when $a=b=c=d=\frac{6}{5}$ and $e=\frac{16}{5},$ so the maximum value of $e$ is $\boxed{\frac{16}{5}}.$