Short answer: yes, because $E_8$ is the unique eight-dimensional lattice with kissing number 240 (the largest possible). (Reference: Theorem 8, Chapter 14 by Bannai and Sloane, from Conway and Sloane, Sphere packings, lattices and groups, 1988.)
(Edit to add: your construction is an instance of what Sloane calls "Construction A for complex lattices", from Section 8, Chapter 7 of Conway and Sloane, whereby you can combine a length $n$ $q$-ary code with an index $q$ ideal of $\Bbb{E}$ or similar to form a complex $n$-dimensional lattice. Example 11b notes that the Tetracode yields $E_8$.)
If you want an explicit isomorphism, then write elements of $E_8'$ as
$v=(a,b,c,d,e,f,g,h)\in\Bbb{Z}^8$, representing
$(a+b\omega,c+d\omega,e+f\omega,g+h\omega)\in\Bbb{E}^4$. The triality vector
$(c(a+b\omega),c(c+d\omega),c(e+f\omega),c(g+h\omega))$ is then $(a+b,c+d,e+f,g+h)\in
T$, taking the co-ordinates modulo 3. Define
$$
M=\begin{pmatrix}
\begin{array}{rrrrrrrr}
0 & 0 & 0 & 0 & 3 & 0 & 3 & 0 \\
0 & 0 & -2 & -2 & 1 & -2 & 1 & -2 \\
0 & 0 & -2 & 4 & 1 & -2 & 1 & -2 \\
0 & 0 & 4 & -2 & 1 & -2 & 1 & -2 \\
0 & 0 & 0 & 0 & 3 & 0 & -3 & 0 \\
-2 & -2 & 0 & 0 & 1 & -2 & -1 & 2 \\
-2 & 4 & 0 & 0 & 1 & -2 & -1 & 2 \\
4 & -2 & 0 & 0 & 1 & -2 & -1 & 2 \\
\end{array}
\end{pmatrix}
$$
Then, considering an element of $E_8'$ as a column vector in $\Bbb{Z}^8$ as described above, multiplying on the left by $\frac16 M$ will give an element of the standard version of $E_8$:
$$\Gamma_8 = \left\{(x_i) \in \mathbb Z^8 \cup (\mathbb Z + \tfrac{1}{2})^8 : {\textstyle\sum_i} x_i \equiv 0\!\!\pmod 2\right\}.$$
Of course this is one of many possible maps since $E_8$ has a large symmetry group, but I
don't think there is a much nicer form of the matrix. Maybe it is not so surprising that
$M$ will look a bit ragged, since the definition of the Tetracode used a particular
choice of (non-symmetric) basis and $M$ has to depend on this choice.
To check this works, we first need to see that $x=\frac16 Mv\in E_8$ for $v\in E_8'$. First
check that each component of $Mv$ is congruent to zero mod 3. Note that pairs of
entries in $M$ satisfy $M_{2i,j}\equiv M_{2i+1,j}\pmod 3$, which means that
$$6x_i\equiv M_{i,0}.c(a+b\omega)+M_{i,2}.c(c+d\omega)+M_{i,4}.c(e+f\omega)+M_{i,6}.c(g+h\omega)\pmod 3.$$
Modulo 3, the only values of $(M_{i,0},M_{i,2},M_{i,4},M_{i,6})$ that arise are
$(0,0,0,0)$, $(0,1,1,1)$, and $(1,0,1,2)$. Dotting these with a Tetracode vector,
$(c(a+b\omega),c(c+d\omega),c(e+f\omega),c(g+h\omega))$, gives zero modulo 3 (easily
checked; also follows from the fact that the Tetracode is self-dual). Thus
$6x_i\equiv0\pmod3$, and so $x_i\in\frac12\Bbb{Z}$.
To see $x_i\equiv x_j\pmod1$, note that each column of $M$ is constant modulo 2. Thus
$6x_i\equiv 6x_j\pmod2$, which is what we want since $6x_i$ and $6x_j$ are also zero modulo 3.
To complete the proof that $x\in E_8$, add the rows of $M$ to see that $$\sum_i x_i=\frac16 \sum_{ij} M_{ij}v_j=2(e-f)\equiv0\pmod2.$$
Finally, to show that the image $\frac16 M(E_8')$ is the whole of $E_8$, it's sufficient to
check that we can reach a basis of $E_8$.
$$\frac16 M\begin{pmatrix}
\begin{array}{rrrrrrrr}
0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 \\
-1 & -1 & 1 & 1 & -1 & 0 & 0 & 0 \\
-1 & -1 & 2 & -1 & 0 & 0 & 0 & 0 \\
1 & -1 & 0 & 0 & 1 & -1 & 0 & 1 \\
0 & -1 & 0 & 0 & 1 & -1 & 0 & -1 \\
1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 \\
\end{array}
\end{pmatrix}=
\begin{pmatrix}
\begin{array}{rrrrrrrr}
1 & -1 & 0 & 0 & 0 & 0 & 0 & 1/2 \\
1 & 1 & -1 & 0 & 0 & 0 & 0 & 1/2 \\
0 & 0 & 1 & -1 & 0 & 0 & 0 & 1/2 \\
0 & 0 & 0 & 1 & -1 & 0 & 0 & 1/2 \\
0 & 0 & 0 & 0 & 1 & -1 & 0 & 1/2 \\
0 & 0 & 0 & 0 & 0 & 1 & -1 & 1/2 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1/2 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1/2 \\
\end{array}
\end{pmatrix}
$$
The columns of the matrix on the left are all in $E_8'$, having triality vectors
$(0,1,1,1)$, $(0,1,1,1)$, $(0,0,0,0)$, $(0,0,0,0)$, $(0,2,2,2)$, $(1,0,1,2)$,
$(0,0,0,0)$, $(0,0,0,0)$ respectively. The columns of the matrix on the right form a
basis of $E_8$.
