If a quadratic form in $n$ variables is the sum of the squares of
$n$ integer linear forms, it's the sum of the squares of $n$ rational linear forms.
Thus it's equivalent
as a rational quadratic form to $x_1^2+\cdots+x_n^2$. In particular
its discriminant is a square. This rules out $E_6$ and $E_7$.
The case of $E_8$ is trickier, as it is equivalent over the rationals
to $x_1^2+\cdots+x_8^2$. This time you have to show there's no
equivalence over the integers, but one form takes solely even values
and the other doesn't.
Added
The first time round I didn't clock the $\ge n$ condition. But there's certainly
a way to put an upper bound on the number $m$ of linear forms one needs.
I'll stick to the $E_8$ form. One can think of this as describing a lattice
$L$ in Euclidean space. This lattice $L$ is self-dual, unimodular and even.
Its shortest vectors form the set of 240 roots $R$. This set of roots
has the nice property (I think it may be called the eutactic property
or the perfect property; all this is in Martinet's book on lattices)
that $x\mapsto\sum_{y\in R}(x\cdot y)^2$ is proportional to the quadratic
form $x\mapsto x\cdot x$. I think actually $\sum_{y\in R}(x\cdot y)^2=30x\cdot x$
as $30=240/8$.
Now an integer linear form is a linear form taking integer
values on the lattice $L$ and so is $x\mapsto x\cdot z$ for some $z$ in the dual
of $L$, so here $z\in L$. If $x\cdot x=\sum_{j=1}^m(x\cdot z_j)^2$
for $z_j\in L$ then
$$480=2|R|=\sum_{y\in R}y\cdot y=\sum_{j=1}^m\sum_{y\in R}(y\cdot z_j)^2
=30\sum_{j=1}^m z_j\cdot z_j.$$
Each $z_j\cdot z_j\ge 2$ so we must have $m=8$ and each $z_j\cdot z_j=2$.
You should check my numbers... For $E_6/E_7$ the dual lattice is different
from the original but I'm sure they still have the eutactic(?) property. Any way
one can get an effective upper bound on $m$, probably not much bigger than $n$.
