It is a general fact that every intermediate model $M\models\text{ZFC}$ sitting between a model and a forcing extension $V\subset M\subset V[G]$ is itself a forcing extension. Indeed, if $G\subset\mathbb{B}$ is $V$-generic for the complete Boolean algebra $\mathbb{B}$, then there is a complete subalgebra $\mathbb{B}_0\subset\mathbb{B}$ such that $M=V[G\cap \mathbb{B}_0]$. (This is proved in Jech's book.)
Thus, if $z$ is any real or indeed any set of ordinals in a forcing extension $V[G]$, then we may form the model $V[z]$, which is a model of ZFC, and by the general fact above it is obtained by forcing with a complete subalgebra of the forcing used with $G$.
In your case, every real $z\in V_1$ has a hereditarily countable name, and thus $z\in V[g_0]$ for some $V$-generic Cohen real in $V_1$. Since every nontrivial subalgebra of $\text{Add}(\omega,1)$ has a countable dense set, it follows that $V[z]=V[g_0]$ for some Cohen real $g_0\in V_1$. In short, every real in $V_1$ that is not in $V$, and this means all $\aleph_{\omega+1}$ of them, is added by the forcing to add a single Cohen real. (These reals are not themselves Cohen reals, but they have names for the forcing to add a Cohen real which interpret to them by some actual Cohen real in $V_1$.)
But I'm not sure if I have understood your question in the sense that you may have intended it. In particular, if you meant that you wanted a single forcing notion to add an entire $\aleph_{\omega+1}$ sequence of reals, then I would say that you already have it, namely, the forcing $\text{Add}(\omega,\aleph_\omega)$ itself was already adding all those extra reals.
Edit. Let me answer Goldstern's follow-up question in the
comments, a question I find very interesting.
First, I claim that the $V[G]=V[H]$ situation he mentions is
impossible, where $G$ is $V$-generic for adding $\aleph_\omega$
many Cohen reals and $H$ is $V$-generic for adding
$\aleph_{\omega+1}$ many Cohen reals. The reason is that if these
two extensions were equal, then it would follow that the two
forcing notions $\text{Add}(\omega,\aleph_\omega)$ and
$\text{Add}(\omega,\aleph_{\omega+1})$ are isomorphic below
respective conditions and hence simply isomorphic. But the former
has a dense set of size $\aleph_\omega$ and the latter has dense
sets only of size at least $\aleph_{\omega+1}$, since any smaller
set than this could not mention enough points in the domains of
the conditions to be dense.
Second, a similar argument shows now that we cannot even have that $V[G]$
contains a $V$-generic filter $H$ for
$\text{Add}(\omega,\aleph_{\omega+1})$, for in this case we would
have that $\text{Add}(\omega,\aleph_{\omega+1})$ is isomorphic to
a complete subalgebra of $\text{Add}(\omega,\aleph_\omega)$. But
since this latter Boolean algebra is $\aleph_\omega$-dense, it
follows that every complete subalgebra of it is also
(at most) $\aleph_\omega$-dense. But $\text{Add}(\omega,\aleph_{\omega+1})$
has no dense set of size less than $\aleph_{\omega+1}$, for the reason described in the previous paragraph. This is a
contradiction, and so $V[G]$ contains no $V$-generic filter for
$\text{Add}(\omega,\aleph_{\omega+1})$.
More generally, essentially the same argument shows that adding $\theta$ many Cohen reals can never add a generic filter for adding $\lambda$ many Cohen reals, if $\theta\lt\lambda$.
But meanwhile, as my answer at the other question shows, $V[G]$ has a family of $\aleph_{\omega+1}$ many pairwise (and finitely-wise) mutually generic Cohen reals. But these do not rise to the level of mutual genericity necessary to form a generic for $\text{Add}(\omega,\aleph_{\omega+1})$.
