Yes, insisting upon a normed vector space rather than a metric vector space introduces important additional properties.
For instance, $\mathbb{R}^n$ is always complete when equipped with a norm, because every norm on a finite-dimensional real (complex) vector space is equivalent, and $\mathbb{R}^n$ has at least one norm under which it is complete, namely the Euclidean norm.
On the other hand, there are metrics under which $\mathbb{R}^n$ is not complete. One example is covered in Why $\mathbb R$ is not complete with the metric $d(x,y)=|\phi(x)-\phi(y)|$ where $\phi(x)=\frac{x}{1+|x|}$?.
There are also complete metrics on $\mathbb{R}^n$ that are not norms; the discrete metric is one easy example.
Therefore if you want to study $\mathbb{R}^n$ as a metric space only, then you must give up on results that depend critically on a norm structure. If you allow yourself incomplete metrics, then you cannot have a norm; and even if you give yourself completeness, a norm is not guaranteed.
Here is also an example of a theorem for finite-dimensional real vector spaces that works for norms but not quite for metrics: Are all the finite dimensional vector spaces with a metric isometric to $\mathbb R^n$. And it is quite easy to see that not all metrics on $\mathbb{R}^n$ are equivalent; the discrete metric cannot be metric-equivalent to any norm. Since a norm scales and the discrete metric does not, no constant $C$ can satisfy $ n(x,y) \leq C d(x,y)$ for all $x,y$, where $d$ is the discrete metric and $n$ is the norm. (Take $x\neq y$ so that the RHS is $C$, then look at $\alpha x,\alpha y$ for $\alpha\to +\infty$.)