Q:

Why study real vector spaces of dimension $n$ other than $\Bbb{R}^{n}$? 

Every vector space of dimension $n$ is isomorphic to $\mathbb{R}^{n}$. Why do we study other (finite-dimensional, real) vector spaces? 

A:

This is an example I use to show what's the importance of avoiding a particular identification of a vector space with $\mathbb{R}^n$ or $\mathbb{C}^n$.
Consider two similar $n\times n$ matrices $A$ and $B$, so there is an invertible matrix $S$ such that $A=SBS^{-1}$ or, that's the same, $AS=SB$.
Let $\lambda$ be an eigenvalue of $A$.  Then we know that $\lambda$ is also an eigenvalue of $B$, because the two matrices have the same characteristic polynomial.  Define
$$
E_A(\lambda)=\{v:Av=\lambda v\}
$$
(in $\mathbb{R}^n$ or $\mathbb{C}^n$ depending on the context).  We want to show that $E_A(\lambda)$ and $E_B(\lambda)$ have the same dimension. 
Consider $w\in E_B(\lambda)$, so $Bw=\lambda w$; then
$$
\lambda Sw=S(\lambda w)=SBw=ASW
$$
which implies $Sw\in E_A(\lambda)$. Similarly, if $v\in E_A(\lambda)$, then $S^{-1}v\in E_B(\lambda)$.
The maps $f\colon E_A(\lambda)\to E_B(\lambda)$, $f(v)=S^{-1}v$, and $g\colon E_B(\lambda)\to E_A(\lambda)$, $g(w)=Sw$, are linear and inverse to one another.  Therefore, they are bijective and the rank-nullity theorem implies
$$
\dim E_A(\lambda)=\dim E_B(\lambda)
$$
Note that even if $E_A(\lambda)$ and $E_B(\lambda)$ are subspaces of $\mathbb{R}^n$ (or $\mathbb{C}^n$) no basis for them has been chosen and that a proof relying on bases would be just the same, only with additional unnecessary complications. 

A:

I understand the question to be, "Every (finite-dimensional, real) vector space is isomorphic to $\mathbf{R}^{n}$, so why study other (finite-dimensional, real) vector spaces?" 
There are many good answers (families of spaces, dual spaces, change of basis, ...).  Here's an example illustrating that vector spaces "with no (natural) basis" arise naturally. 
Let $S$ denote the unit sphere in Euclidean three-space.  For each point $p$ of $S$, there is a tangent plane $T_p S$. This plane is a two-dimensional vector space, but comes equipped with no natural basis. ( Just because a tangent plane $T_{p}S$ is isomorphic to $\mathbf{R}^{2}$ does not mean $T_{p}S$ is $\mathbf{R}^{2}$.)
Moreover, "the family of all tangent planes to $S$ admits no continuous choice of basis".  In more conventional language, there do not exist two continuous, pointwise linearly independent tangent vector fields on $S$, since there does not even exist a continuous, non-vanishing tangent vector field on $S$.
That is, even surface geometry in ordinary space forces you to study vector spaces that do not come equipped with a basis, i.e., that are not $\mathbf{R}^{2}$.

