Hi Noah,

Consider the forgetful morphism $(\mathbb P^d)^{n+1}\to (\mathbb
P^d)^n$. This restricts to a forgetful morphism $\pi:V_{d,n+1}\to V_{d,n}$. This is
kind of like the map $\overline{\mathscr M}_{g,n+1}\to \overline{\mathscr
M}_{g,n}$. The general fiber of $\pi$ is a $\mathbb P^1$ (the original degree $d$
rational normal curve).

Now let's see normality. Usually the best way to prove normality is via Serre's
criterion: normal is equivalent to $R_1$ and $S_2$. So, you want to prove these
conditions.

For $n=d+3$ it is easy since $V_{d,n}=(\mathbb P^d)^n$. So it is normal (actually smooth of
course). In particular it is both $R_1$ and $S_2$, in fact it is Cohen-Macaulay
(CM).

Now consider $\pi:V_{d,n+1}\to V_{d,n}$. If $V_{d,n}$ is $R_1$, then I think
(likely) so is $V_{d,n+1}$: the singular set of $V_{d,n+1}$ is contained in the union
of the pre-image of the singular set of $V_{d,n}$ and the locus where $\pi$ is not
smooth.
The former is of at least codimension $2$ by induction. I am not entirely certain
about the latter, but in any case the non-smooth locus has two parts: the locus of
fibers with multiple components and the locus of singular set of the reduced singular
fibers.
The latter of these is definitely of at least codimension $2$ since the locus of
these fibers is of at least codimension $1$ and the singular part is at least
codimension $1$ in that. So, you need that the locus of non-reduced fibers is at
least codimension $2$. I think this seems OK, but this is one of those things I did
not carefully work out. It seems to me that (any component of this) ought to be
contained in (a component of) the reduced singular locus. The reason I think that is
that it seems that there should be a partial "smoothing" of a non-reduced fiber when
you just separate the components. If this is true, then the locus is at least of
codimension $2$. Then again, if this fails then normality fails, so at least you got a necessary condition.

This should take care of condition $R_1$. Now on to $S_2$.

In some sense this is harder, because we need this condtion at the singular set as
well while above we just needed to show that the singular set is not too big.

For $d=2$ this is easy, because a conic (including degenerate ones) is defined by a single equation. So we can
prove that $V_{d,n}$ is actually CM (=Cohen-Macaulay) and hence $S_2$. This follows
by induction: If $V_{d,n}$ is CM, then so is $V_{d,n}\times \mathbb P^d$ and
$V_{d,n+1}$ is a Cartier divisor in $V_{d,n}\times \mathbb P^d$ and hence itself is
CM.

The same proof does not work for $d>2$. Although a smooth curve in a smooth total space is a local complete intersection, as pointed out by mdeland in the remarks this does not remain true for all degenerations. (I should have realized this as this is a famous example...)
At the same time I feel that one might still be able to do something like this. After all we do not need the individual curves to be lci or even S_2. (If we did, this example would lead to non-normality).

It seems this definitely works for $d=2$ but not for other $d$'s. However, at least this might give you some ideas on how one might go about proving something like this. I would add
that if this does not turn out to be normal, then perhaps this is not the "right"
compactification to consider. At the least you should assume that your degenrate curves have no embedded points, but I could imagine that the best to do is to demand that degenerations be *stable*, i.e., consider the pointed curve degenerating to a pointed degenerate curve
that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above
proof would essentially prove that it is normal. Of course, this may be what you're
doing in general or it does not work for some reason, I am not familiar with the
relevant results, so this is just the first thing I would try.