On the definition of the volume form in general vector spaces as given in Spivak, Calculus on Manifolds For a vector space $V$ denote by $\Lambda^k(V)$ the space of alternating $k$-tensors, or alternating $k$-fold multilinear maps on $V$. I have some difficulty following the intention of the author in the part where he introduced the volume form of a finite-dimensional vector space $V$ (page 83):

The fact that $\dim \Lambda^n(\mathbb R^n) = 1$ is probably not new to you, since $\det$ is often defined as the unique element $\omega \in \Lambda^n(\mathbb R^n)$ such that $\omega(e_1, \ldots, e_n) = 1$. For a general vector space $V$ there is no extra criterion of this sort to distinguish a particular $\omega \in \Lambda^n(V)$. Suppose, however, that an inner product $T$ for $V$ is given. If $v_1, \ldots, v_n$ and $w_1, \ldots, w_n$ are two bases which are orthonormal with respect to $T$, and the matrix $A = (a_{ij})$ is defined by $w_i = \sum_{j=1}^n a_{ij} v_j$, then
  $$
 \delta_{ij} = T(w_i, w_j) = \sum_{k,l = 1}^n a_{ik}a_{jl} T(v_k, v_j) = \sum_{k=1}^n a_{ik}a_{jk}.
$$
  In other words, if $A^T$ denotes the transpose of the matrix $A$, then we have $A \cdot A^T = I$, so $\det A = \pm 1$. It folows as $\omega(w_1,\ldots, w_n) = \det(A)\cdot \omega(v_1, \ldots, v_n)$ that if $\omega \in \Lambda^n(V)$ satisfies $\omega(v_1, \ldots, v_n) \in \{-1,1\}$, then $\omega(w_1, \ldots, w_n) \in \{-1,1\}$. If an orientation $\mu$ for $V$ has also been given, it follows that there is a unique $\omega \in \Lambda^n(V)$ such that $\omega(v_1, \ldots, v_n) = 1$ whenever $v_1, \ldots, v_n$ is an orthonormal basis such that $[v_1,\ldots, v_n] = \mu$. This unique $\omega$ is called the volume element of $V$.

First why he claims that for a general (finite-dimensional) $V$ there is no extra criterion as for $\mathbb R^n$; if we choose some basis $v_1, \ldots, v_n$ of $V$ then the requirement $\omega(v_1, \ldots, v_n) = 1$ gives me a unique $n$-form (as the value on every permutation of the basis vectors is then also prescribed). Of course, this choice is relative to a basis, but that was the choice in $\mathbb R^n$ also. And the orientation is coded in the basis. So why all these complicated derivations using the given orientation and an inner product, when simply a basis is enough?
Also if we have a basis of a finite-dimensional vector space, we also get an inner product by multiplying the coefficient w.r.t. this basis and summing. Of course the geometric meaning of such an inner product is questionable, as it no longer corresponds to angles, length and projections as in the $\mathbb R^n$ setting with the standard inner product if the $v_i$ are not orthonormal. But for mere existence of an inner product such that the basis is orthonormal w.r.t. it, it is enough.
I can image that such definitions are not very meaningful from a geometrical point of view, if we take an arbitrary basis then built our volume form by the mere requirement that it takes the value one on this basis, we have no ready-to-use interpretation of it as an actual volume, as for it we need orthonormality and a notion of length, angle and so on already given by a prescribed inner product. (of course by the above construction of inner product an angle of two vectors w.r.t. to this constructed inner product corresponds to the "real" geometric angle if we view the coordinates as coordinates of the standard basis in $\mathbb R^n$). So I guess that what he might want to say is "there is no sort to distinguish any particular form that has also any meaningful geometrical interpretation, as for example given by some inner product and orientation, after which we can talk about (oriented) length, areas, volumes". But this is just a guess of mine; as how we are interested in "real world interpretations" in formal mathematics and its definition is here not clear to me.
Maybe what he want to say is that the essential ingredients for a volume form (as derived from $\mathbb R^3$ and its interpretation of volume) is that we need an orthonormal basis (hence an inner product), which also fixes an orientation, such that a meaningful volume form could be introduced. 
 A: Yes, it's true that a choice of basis determines an element of $\Lambda^n(V)$. But that's not very useful for manifolds, because there's rarely a natural way to single out a particular basis of each tangent space on a smooth manifold. 
The point is that you don't need anything as specific as a choice of basis to determine a unique $n$-form. For example, if $V$ is endowed with an inner product and an orientation, then it turns out that every oriented orthonormal basis determines the same $n$-form, so in fact the $n$-form is completely determined by the inner product and the orientation.
For manifolds, this is fortunate, because on an oriented Riemannian manifold an inner product and an orientation are exactly the distinguished information we have for each tangent space. Thus every oriented Riemannian manifold has a uniquely determined volume form.
This is not the only way to determine a unique volume form. For example, a nondegenerate 2-form on a vector space determines a volume form, so every symplectic manifold has a uniquely determined volume form. Even more simply, if $M$ is parallelizable (e.g., any Lie group), then every choice of global frame determines a volume form (because it determines a choice of basis at each point).
A: To make my comment better understandable: what is lurking in the background here is the Riesz representation theorem, which says that, after the choice of a scalar product on a vector space V,  there is a unique isomorphism between $V$ and $V^*$ given by $v\mapsto (w\mapsto \langle v, w\rangle)$. 
While it is known that $V$ and $V^*$ are isomorphic, there is normally no natural choice for an isomorphism (you have to choose a basis, but this is arbitrary), but there is one once you have a scalar product. 
So, while it is no surprise that the introduction of a scalar product allows you to define length (and, as a consequence, volume) and while it is also no surprise that $V$ and $V^*$ are isomorphic (in finite dimensions), it should come as a surprise to the naive observer that volume can be (naturally) calculated using elements from $V^*$ once you define length and angles for $V$.
Regarding your question to my comment: the calculation you cite from Spivak shows that with a scalar product the volume form does not depend on the basis, but remains invariant under transformations which preseve the scalar product. So the volume form depends only on the choice of scalar product (and orientation, of course), as mentioned before.
(Sorry, too long for a comment, not intended to answer all of your questions, only the first one). 
