Where does the really nice '8-dimensional' description of the $E_7$ root system come from?

I don't know who found this presentation first but I can imagine that already Cartan knew it since it comes from a symmetric space. More precisely, $\mathfrak g=E_7$ has an involution $\theta$ whose fixed point set is $\mathfrak k=\mathfrak{sl}(8)$. The $-1$-eigenspace of $\theta$, customarily denoted by $\mathfrak p$, is a representation of $\mathfrak k$. In fact, it is the fourth fundamental representation $\wedge^4\mathbb C^8$. So we have a decomposition $E_7=\mathfrak{sl}(8)\oplus\wedge^4\mathbb C^8$. Now your presentation is clear: the permutations of $(1,-1,0,0,0,0,0,0)$ are the root vectors of $\mathfrak{sl}(8)$ while the vectors $(\pm1/2,\ldots\pm1/2)$ are the weights of $\wedge^4\mathbb C^8$.

Similar games can be played with all symmetric spaces. For example $E_8=\mathfrak{so}(16)\oplus\mathbb C^{128}$ (the latter is the spin representation) is a very popular choice. This leads to the root vectors all permutations of $(\pm1,\pm1,0,0,0,0,0,0)$ and all $(\pm1/2,\ldots\pm1/2)$ with an even number of minus signs.

[edited to include more examples of the construction and fix a typo]

This can also be seen purely in terms of lattices or quadratic forms. As you note, for any $n$ the vectors in the slice $\sum_{i=0}^n x_i = 0$ of ${\bf Z}^{n+1}$ constitute the root lattice $A_n$; so we'll use the case $n=7$ of the following general picture.

The dual lattice $A_n^*$ consists of the intersection of this slice with the sublattice of $(n+1)^{-1} {\bf Z}^{n+1}$ with all $x_i \equiv x_j \bmod\bf Z$; the map taking $(x_0,\ldots,x_n) \in A_n^*$ to the common value of $((n+1) x_i) \bmod n+1$ descends to an isomorphism $A_n^*/A_n \cong {\bf Z}/(n+1){\bf Z}$. Any intermediate lattice then consists of the intersection of $A_n^*$ with $d^{-1} {\bf Z}^{n+1}$ for some factor $d$ of $n+1$; this lattice is integral iff $d^2 | n+1$, in which case it is even unless $n$ and $(n+1)/d^2$ are both odd. In particular, taking $n=7$ and $d=2$ we obtain an even lattice of discriminant $8/2^2 = 2$ that contains $A_7$ with index $2$; this must be the lattice $E_7$ (for instance because we can count ${8 \choose 4} = 70$ new roots which together with the $8 \cdot 7 = 56$ roots of $A_7$ give a total of $126$).

The root lattice $E_8$ can be constructed similarly with $(n,d) = (8,3)$ as a lattice intermediate between $A_8^{\phantom*}$ and $A_8^*$, with index $3$ in both directions.

Other examples: $(n,d) = (2,3)$ gives a lattice isometric with ${\bf Z}^3$; for $(n,d) = (15,4)$ we get the unique unimodular lattice of rank $15$ with no vectors of norm $1$; for $(17,3)$, one of the two indecomposable even lattices of rank $17$ and discriminant $2$ (the other one has the same theta series, and contains $D_{10} \oplus E_7$ with index $2$); and for $(24,5)$, the Niemeier lattice with root system $A_{24}$.

(If you like coding theory or the Fano plane, you can also use a symmetrical picture of $E_7$ in ${\bf R}^7$ that consists of all vectors $2^{-1/2} a$ with $a \in {\bf Z}^7$ such that $a \bmod 2$ is an element of the $[7,3,4]$ code, or equivalently either zero or the complement of a line in the Fano plane. NB $[7,3,4]$ is the dual Hamming code; using the Hamming code $[7,4,3]$ itself yields $E_7^*$, while the extended [8,4,4] Hamming code yields $E_8$.)

The span of the roots of E7 is the pentacontihexapentacosiheptacontihexaexon, aka Gosset polytope 2_13 = x3o3o3o *c3o3o3o. That one allows for a complemental compound decomposition into the small petated hexadecaexon = expanded octaexon = x3o3o3o3o3o3x and the hexadecaexon = trirectified octaexon = o3o3o3x3o3o3o.

In fact, the first one has 56 vertices, the latter one has 70 vertices. If these are compounded in the given orientation wrt. A7 (which most symmetrically can be represented within cartesian coordinates of one dimension plus), then the convex hull of these 126 vertices is nothing than that Gosset polytope 2_13.

You'll find that also on https://bendwavy.org/klitzing/incmats/laq.htm