One approach is to calculate the orbits of $W(A_8)$ on $W(E_8) / W(A_8)$

I claim these orbits have sizes  $1, 1, 84, 84, 560, 560, 630$.

Given this claim, it's straightforward to check. For any intermediate subgroup $G$, $G / W(A_8)$ must have an order a divisor of $1920$ and must be a union of these orbits, including $1$. We can't write $1920/2= 960$ as a sum of these numbers because $630+1+1+84+84$ is too small but including two of $560, 560,630$ would be two big. We can't write $1920/3=640$ for similar reasons. Because $1920/4 =480<560$, the only remaining possibilities are $1, 2, 85, 86, 169, 170$ and none of those is a divisor of $1920$ except $1$ and $2$, which correspond to $W(A_8)$ and its normalizer. 

The way I calculated this involves observing that because the inclusion $W(A_8) \subset W(E_8)$ comes from viewing the lattice $A_8$ as an index $3$ sublattice of $E_8$, the kernel of a map $E_8 \to \mathbb Z/3$ arising by dot product with an element of $E_8$ mod $3$, we can represent $W(E_8)/W(A_8)$ as a $W(E_8)$-orbit inside $E_8$ mod $3$ - specifically, the elements with norm congruent to $2$ mod $3$ that aren't roots. 

I then found all these elements in a model of $E_8$ on which $W(A_8)=S_9$ acts, i.e. vectors in $((1/3) \mathbb Z)^9$ that sum to zero, and calculated the $S_9$ orbits. They are

$630$ permutations of $(0,1,1,1,1,-1,-1,-1,-1)$

$1$ permutation of $(1/3,1/3,1/3,1/3,1/3,1/3,1/3,1/3,1/3)$ (after subtracting $3$ from one entry - mod $3E_8$, it doesn't matter which one)


$84$ permutations of $(1/3,1/3,1/3, -2/3,-2/3,-2/3,-2/3, -2/3, -2/3)$  (after adding $3$ to one entry)

$560$ permutations of $(4/3, 4/3, 4/3,-2/3,-2/3,-2/3,1/3,1/3, 1/3) $ (after subtracting $3$ from one entry)

and the negations of the last three orbits.