Does assigning a different inner product to a vector space in $\mathbb{R^n}$ change the meaning of the determinant on that space? We just started talking about inner product spaces and and how one can assign a different notion of length and angle on a vector space. Since the determinant in $\mathbb{R^n}$ captures the notion of "$n$-dimensional oriented volume", does this mean that a different notion of length will give a different meaning to the determinant? 
Is the "length" defined by an inner product even connected to the "volume" defined by the determinant?
 A: Consider a local isometry $\varphi: \mathbb R^n \to M$.  This induces a new metric whose properties can be seen as follows.  Let $\underline \varphi$ be the differential ("Jacobian matrix")of $\varphi$, and let $\overline \varphi$ be the differential's adjoint ("transpose").
Tangent vectors in $M$ can then be pulled back to $\mathbb R^n$ by $\underline \varphi^{-1}$, and cotangent vectors by $\overline \varphi$.  This allows us to use the Euclidean metric of $\mathbb R^n$ to compute lengths and distances in $M$.
Say you have a curve $c(t): I \to M$.  You find the length of this curve by an integral:
$$\ell = \int_a^b \sqrt{ \left (\underline \varphi^{-1}[c'(t)] \cdot \underline \varphi^{-1}[c'(t)]\right)} \, dt$$
You pull back the tangent vectors to $\mathbb R^n$ and then find the lengths as usual.
Now, what we can do to talk about areas and such is define a wedge product of tangent or cotangent vectors, one that is antisymmetric on exchange.  Let $d(u,v)$ trace out a surface in $M$, then $\partial_u d$ and $\partial_v d$ are vectors, and we can denote an infinitesimal area element by $\partial_u d \wedge \partial_v d \, du \, dv$.  Still, we have to have a "dot product" of this object with itself to get to a scalar, and we have to square root that to get to an area (I won't explain how this dot product is defined, but it does exist and has a concrete definition).
$$A = \int \int \sqrt{ \left (\underline \varphi^{-1}[\partial_u d \wedge \partial_v d] \cdot \underline \varphi^{-1}[\partial_u d \wedge \partial_v d ] \right)} \, du \, dv$$
When we get to $n$-volumes, it's much the same story.  But here's where things get interesting:  it turns out that $\underline \varphi^{-1}(v_1 \wedge \ldots \wedge v_n) = v_1 \wedge \ldots \wedge v_n/\alpha$, where $\alpha$ is some scalar.  This $\alpha$ is by definition the determinant of $\underline \varphi$, so that an $n$ volume integral becomes
$$V_n = \int [\det \underline \varphi]^{-1} \sqrt{(v_1 \wedge \ldots \wedge v_n)^2} \, d^n x$$
Often we write instead that this is the square root of the metric for general spaces.  You can verify that the metric on $\underline g'$ on $M$ depends on $\overline \varphi^{-1} \underline \varphi^{-1}$, so $\det \underline g' = [\det \underline \varphi]^{-2}$.
So a different notion of length--a different metric--will naturally lead to different measurements for volumes, but the volume is still related to the determinant of the metric in every space.  It's only in cartesian coordinates for Euclidean space, where the metric is the identity, that taking the determinant of the vectors works out the same as doing the integral above, with the determinant of the metric appearing as well.
The metric determines lengths, as it defines the inner product, and measures of the volume depend on it as well.
A: I think this interpretation of the determinant as "volume" only holds for positive definite metrics over $\Bbb R$ spaces. If you change the units of measure on the axes, the determinant will accomodate this by changing accordingly. (I mean, if you first measured in meters and got a volume of 1 cubic meter, making a coordinate change to centimeters would change the determinant to $100^3$ centimeters.)
If for some strange reason you measured feet on one axis, inches on another and meters on the last, then you would still get the volume measurement out of the determinant, it's just that it would be in the rather strange units: feet*inch*meters.
In general, the determinant of a matrix does not use any inner product in its computation. You can even do it over fields for which ordering, length and volume really does not make (Euclidean) sense. It's probably better to think of the determinant being attached to a linear transformation. You probably know that all matrix representations of a given transformation are conjugate, and since the determinant is conjugation invariant, you can see that the determinant is really more attached to the transformation rather than the matrix.
A: To continue @rschwieb's good answer: indeed, first, "volume" probably only refers to real vector spaces, and there must be some measure (if not in a terribly technical sense) attached, too, or we don't know how to measure [sic] volume. That is, an "abstract" $n$-dimensional $\mathbb R$-vectorspace $V$ has no canonical volume. It is completely ambiguous.
Yet, as rschwieb noted already, the notion of determinant of a linear transformation on $V$ does not use a volume or basis: $\det A$ is the scalar by which $A$ acts on the highest non-vanishing exterior power of $V$. The highest non-vanishing exterior power is one-dimensional, so its endomorphisms are scalars... 
Thus, the real question is about how we know that, or why, determinants do compute volume in the familiar case. One can discern that the "whole" of the inner product is not needed, because a stretch in one coordinate axis can be compensated-for by a shrink in another, as rschwieb noted.
In my opinion, the most direct understanding is that an invertible element of the group $GL_n(\mathbb R)$ of invertible real matrices has a "Cartan decomposition" (a.k.a. "singular value decomposition") $k_1\delta k_2$ where $k_j$ are orthogonal and $\delta$ is positive-diagonal. "Orthogonal" is metric-preserving, so certainly volume-preserving. Positive-diagonal matrices $\delta$ arguably change volume in a manner predictable from their effect on rectangular boxes: stretching/shrinking by amount $\delta_i$ (the $i$th diagonal entry) changes the volume of a coordinate-axis-aligned box by $\prod_i \delta_i$.
From this, the volume change by $A=k_1\delta k_2$ is $|\det \delta|=\det A$, since determinants of orthogonal matrices are $\pm 1$.
