In your extension $V_1$, there are actually continuum many,
that is $\aleph_{\omega+1}$ many $V$-generic Cohen reals.
In general, whenever you add even a single Cohen real, then
the extension wills have continuum many $V$-generic Cohen
reals, because if $c$ is a $V$-generic Cohen real and $x$
is any real in the ground model, then the bit-wise sum
$c\oplus x$ will be a $V$-generic Cohen real, since this
induces an automorphism of the forcing in the ground model,
and these are all different. (In your case, as Goldstern mentions in the comments, we may view the forcing as adding all but one of the Cohen reals, and then a final Cohen real, to achieve $\aleph_{\omega+1}$ many $V$-generic Cohen reals by this reasoning.)

But it is true that not every new real of $V[G]$ is a
$V$-generic Cohen real. For example, one may easily
construct reals that obey some regular pattern, repeating
their digits in pairs, for example, which prevent them from
being Cohen reals, even if they are not in the ground
model.

Nevertheless, if $V[G]$ is a forcing extension obtained by
adding any number of Cohen reals and $z$ is any real in
$V[G]$, then I claim that $z\in V[c]$ for some $V$-generic
Cohen real in $V[G]$. This is because the countable chain
condition of the forcing means that one needs only
countably much information from $G$ in order to construct
$z$, and restricting the generic sequence of Cohen reals to
any countable domain is isomorphic again to adding a single
Cohen real.

In general, if $V\subset V[G]$ is any forcing extension and
$W$ is a model in between $V\subset W\subset V[G]$, such as
$W=V[z]$ for some real $z$, then $W$ is also a forcing
extension of $V$ by a complete subalgebra of the Boolean
algebra giving rise to $G$.

In the case of adding a Cohen real, forcing which has a
countable dense set, every subalgebra of its Boolean
algebra also has a countable dense set, and all nontrivial
forcing notions with a countable dense set are isomorphic
to adding a Cohen real. In this sense, every real added in
a Cohen real forcing extension (and this includes every
real in your model by my observations above) is generic
over $V$ for forcing that is isomorphic to the forcing to
add a Cohen real.

So the answer to your final question is that yes, these extra reals are $V$-generic for some forcing notion, and that forcing notion is isomorphic to the forcing to add a single Cohen real!

Let me conclude with an interesting tidbit:

**Theorem.** In the forcing extension $V[c]$ obtained by
adding a single $V$-generic Cohen real, there is a family
of continuum many pairwise mutually generic Cohen reals.
Indeed, there is a perfect set $P$ in $V[c]$, all of whose
finite subsets are mutually $V$-generic Cohen reals.

Proof. Consider the forcing to add such a perfect set $P$.
We want to force to create a tree, all of whose branches
are $V$-generic Cohen reals, and such that any finitely
many branches are mutually generic Cohen reals. Let
conditions be finite binary trees, ordered by
end-extension. It is dense for the leaves to be extended
into any given dense subset of Cohen forcing. And for
finite products of Cohen forcing with itself, it is dense
to extend the tree so that all pairs (or triples etc.) of
branches are inside any given dense set in the product
forcing. Thus, this forcing will create such a tree and
hence such a perfect set.

Finally, observe that our tree forcing has only countably
many conditions, and thus it is isomorphic to adding a
single Cohen real. So the forcing extension $V[c]$ already
has such a pefect set. QED

Of course, the branches through the perfect set will not be
$V$-generic for the forcing to add continuum many Cohen
reals, since that forcing is not a subalgebra of the
forcing to add only one. The reals in the perfect set are
only mutually generic when taken finitely many at a time,
but not fully mutually generic for infinite collections.
For example, the perfect set contains reals that are the
limits of other of its elements, and this violates mutual
genericity for those infinite families.