soft question - Alternating forms as skew-symmetric tensors: some inconsistency?

I can't speak to what is actually used, particularly what is used by physicists! However, I can try to shed some light on the diagram and the maps in question. In actual fact, there are two diagrams here and you are conflating them. This, simply put, is the source of the confusion. Let me expand (at a bit more length than I intended!) on that.

Firstly, there are too many maps flying around and some are more canonical than others. The most canonical is the identification of $(bigotimes^k V)^*$ with $operatorname{Mult}^k(V)$ since this is by (one of the) definition(s) of the tensor product. So let us start with that. The inclusion $operatorname{Alt}^k(V) to operatorname{Mult}^k(V)$ is probably next in line since it is the inclusion of a subspace. After that, I'd put the map $bigotimes^k V^* to (bigotimes^k V)^*$. So, so far we have a diagram:

$$
begin{array}{ccccc} operatorname{Alt}^k V \
i downarrow \
operatorname{Mult}^k V &leftarrow & (otimes^k V)^* & leftarrow & otimes^k V^*
end{array}
$$

That the horizontal maps are isomorphisms is nice, but only holds for finite dimensional vector spaces so I'm not going to write in the fact that they are isomorphisms. I want to emphasise what's really canonical and what's not.

Now let us consider $(Lambda^k V)^*$. We appear to have a canonical map from this to $operatorname{Alt}^k(V)$ but in fact, we don't. We have a canonical map from this to $(bigotimes^k V)^*$ given by:

$$
f(v_1 otimes cdots otimes v_k) = f(v_1 wedge cdots wedge v_k)
$$

This is dual to the projection map $bigotimes^k V to Lambda^k V$. That projection map is pretty canonical as we usually define $Lambda^k V$ as a quotient of $bigotimes^k V$. Taking its dual is a natural thing to do, so this also appears on my list of "canonical maps". Now when we go "down" and "across" we happen to end up in the subspace $operatorname{Alt}^k(V)$ so we can add a horizontal arrow $(Lambda^k V)^* to operatorname{Alt}^k(V)$ if we like, but the new map that we add by doing this is one step removed from the really canonical maps so I'm going to leave it out at this stage.

Now we come to $Lambda^k V^*$. This is, as for $Lambda^k V$, defined as a quotient of the tensor product. So we have a projection $bigotimes^k V^* to Lambda^k V^*$. This, again, is pretty canonical. So our "canonical" diagram looks like this:

$$
begin{array}{ccccc} operatorname{Alt}^k V && (Lambda^k V)^* && Lambda^k V^* cr
i downarrow &&{p_V}^* downarrow&& uparrow p_{V^*}cr
operatorname{Mult}^k V &leftarrow & (otimes^k V)^* & leftarrow & otimes^k V^*
end{array}
$$

At this point, an obvious question is as to whether or not we can fill in the gaps. I've already said that we can in the top-left. Can we in the top-right? That is, is there a map $Lambda^k V^* to (Lambda^k V)^*$ making the diagram commute? (Thinking about infinite dimensions says that this is the correct direction.) The answer is: (drum roll) No. And the reason is quite simply that we start in $bigotimes^k V^*$ and can choose any element there as our starting point, but would want to end up in the alternating part of $(bigotimes^k V)^*$.

Okay, now we throw in the Alternator (probably time for another drum roll). The Alternator does what it says on the tin: it alternates stuff. But we have to be careful and ensure that we only apply it to stuff that can genuinely be alternated. So we have an alternator: $operatorname{Alt} colon operatorname{Mult}^k(V) to operatorname{Alt}^k(V)$ given by

$$
operatorname{Alt}(f)(v_1,dotsc,v_k) = frac{1}{k!} sum (-1)^{sigma} f(v_{1sigma}, dotsc, v_{ksigma})
$$

The $1/k!$ is to make this a left inverse of the inclusion $operatorname{Alt}^k(V) to operatorname{Mult}^k(V)$.

We also have an alternator $Lambda^k V to bigotimes^k V$ given by:

$$
v_1 wedge dotsb wedge v_k mapsto frac{1}{k!} sum (-1)^{sigma} v_{1sigma} otimes dotsb v_{ksigma}
$$

Again, the multiplier is chosen to ensure that this is a right inverse of the projection map. This is your $Sk$ map. Putting these into a diagram, we get:

$$
begin{array}{ccccc} operatorname{Alt}^k V && (Lambda^k V)^* && Lambda^k V^* cr
operatorname{Alt} uparrow &&{Sk_V}^* uparrow&& downarrow Sk_{V^*}cr
operatorname{Mult}^k V &leftarrow & (otimes^k V)^* & leftarrow & otimes^k V^*
end{array}
$$

Again, the obvious question is: can we fill in the gaps? We can fill in the first one. Indeed, the same filler map works in this diagram as in the last. That was the map $alpha colon (Lambda^k V)^* to operatorname{Alt}^k(V)$ with the property that $i alpha = eta {p_V}^*$ (where $eta colon (bigotimes^k V)^* to operatorname{Mult}^k(V)$ is the isomorphism). So:

$$
i alpha (Sk_V)^* = eta {p_V}^*(Sk_V)^* = eta (Sk_V p_V)^* = eta;text{and}; i operatorname{Alt} eta = eta
$$

Thus, as $i$ is an injection, $alpha (Sk_V)^* = operatorname{Alt} eta$.

But it's the other gap that's more interesting. Now we can fill it in. And the "filler" map is laid out for us already: it's simply follow-the-arrows. If we work it out in detail, it's the following map:

$$
begin{aligned}
f_1 wedge dotsb wedge f_k mapsto Big((v_1 wedge dotsb wedge v_k) mapsto & Sk_{V^*}(f_1 wedge dotsb wedge f_k) big( {Sk_V}^*(v_1 wedge dotsb wedge v_k)big)Big) \
&= frac{1}{k!} frac{1}{k!} sum_sigma sum_tau (-1)^{sigma} (-1)^{tau} f_{1sigma}(v_{1tau}) dotsb f_{ksigma}(v_{ktau})
end{aligned}
$$

This simplifies considerably by rewriting $f_{jsigma}(v_{jtau})$ as $f_{jrho}(v_j)$. Then we end up with $k!$ of each term, so we get:

$$
(f_1 wedge dotsb wedge f_k)(v_1 wedge dotsb wedge v_k) = frac{1}{k!} operatorname{det}(f_i(v_j))
$$

But notice the factor of $1/k!$ in this!

So to make that right-hand rectangle commute, one of the maps has to have a factor of $1/k!$ in it. It doesn't have to be the top one, but that's the most obvious one since if you modify one of the $Sk$s then you ought to modify the other one - though there's no reason to do so, and in fact this might be what's going on: the physicists are keeping one of the $Sk$s as it is and defining the other one to be suitably scaled so that the upper map is the determinant map. But that's speculation, returning to reality we have a diagram:

$$
begin{array}{ccccc} operatorname{Alt}^k V && (Lambda^k V)^* &stackrel{frac{1}{k!}operatorname{det}}{leftarrow} & Lambda^k V^* cr
operatorname{Alt} uparrow &&{Sk_V}^* uparrow&& downarrow Sk_{V^*}cr
operatorname{Mult}^k V &leftarrow & (otimes^k V)^* & leftarrow & otimes^k V^*
end{array}
$$

Finally, let's compare this to your original diagram. The key thing to notice is that in my diagrams, I have two vertical maps in one direction and one in the other. In your diagram, you have two vertical maps in the same direction (and are missing the third). But whichever of my diagrams you prefer, one of your maps is going in the wrong direction.
So, in conclusion, the mistake is that your diagram isn't supposed to commute. Rather, there's two commuting diagrams there with some maps from one diagram and some from another.

(I have a feeling that I haven't really answered the question. This was what I wrote out when trying to make sense of the question rather than towards an answer. But I hope that it helps clarify the issue for you.)