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.
Whether or not the same proof works for $d>2$, depends on the degenerations of rational normal curves of degree $d$ and you probably know more about these than I do. Anyway, a smooth curve in a smooth total space is a local complete intersection, so the question here is whether this remains true for all degenerations. If it does then $V_{d,n+1}$ is a local complete intersection in $V_{d,n}\times \mathbb P^d$ and then by the above argument itself is CM if the previous one was CM.
I think this definitely works for $d=2$ and possibly for other $d$'s, but as you can tell I did not work out all the details. 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. I could imagine that you 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.