Just saw this thanks to a "Related" link from
Question 154928.
Yes, it is known that the $E_6$ form cannot be written as a sum of
integral squares, and thus (by specialization) that the same is true of
$E_7$ and $E_8$; moreover the representations of $D_n$ ($n>2$) and $A_n$
as the sum of $n$ and $n+1$ squares respectively are the only ways
(up to isomorphism) to write these forms as the sum of any number of
*nonzero* squares, with the exception of $D_3 \cong A_3$ which has
both a three-square and a four-square representation.
Or at least it is known once one makes **Will Jagy**'s
key observation that writing a form as a sum of $m$ integral squares is
tantamount to embedding the corresponding lattice into ${\bf Z}^m$.
(As it happens I was just asked a few days ago whether any integral
positive-definite lattice can be embedded in some ${\bf Z}^m$,
so this question was very familiar.)

I don't know a reference, but the result is not hard starting from the
Coxeter diagrams
and the fact that the vectors of norm $2$ in ${\bf Z}^m$ are exactly
the vectors $e+e'$ for some $\pm$ unit vectors $e,e'$ with $e' \neq \pm e$.

For $A_n$ we need a sequence of $n$ such vectors any two of which
are orthogonal except that consecutive vectors have inner product $-1$.
The first two must be $e'-e, \, e''-e'$ with $e,e',e''$ orthogonal
unit vectors. The third could be either $e'''-e''$ or $e+e'$.
The latter choice does not extend to $A_4$,
and the former extends uniquely to $e''''-e'''$, and then by induction to
$\{ e^{(i)} - e^{(i-1)} \}_{i=1}^n$ with all $n+1$ unit vectors orthogonal.

For $D_n$ we need an $A_{n-1}$ configuration together with a norm-$2$ vector
orthogonal to all but the second vector, with which it has inner product $-1$.
For $n=3$ we've done this already because the $D_3$ and $A_3$ diagrams
are isomorphic. For $n=4$ either of our two $n=3$ solutions extends uniquely
and both give $e'-e, e''-e', e'''-e'', e+e'$. For all $n > 4$ the unique
$A_{n-1}$ diagram extends uniquely, again with extra vector $e+e'$.

We can now obtain the impossibilty of the $E_6$ configuration
by trying to extend either $D_5$ at a short end or $A_5$ at the middle vertex
(or by trying to overlap $D_5$ with $A_5$ or with another $D_5$).
Since $E_7$ and $E_8$ contain $E_6$, they are impossible too.

Hence none of the $E_n$ lattices are contained in any ${\bf Z}^m$,
whence the corresponding quadratic forms are not sums of integral squares,
**QED**.