A Banach space is a complete, normed, vector space.
What is the set of vectors $\mathbb{R}^n$ in terms of a Banach space?
Meaning, is $\mathbb{R}^n$ a Banach space? If so, is it a specific type of Banach space? If not, what assumptions can be added to make $\mathbb{R}^n$ a Banach space?
On
To be a Banach space the vector space must be equipped with a norm. When you speak of $\mathbb R^n$ you only have a vector space. So, no it is not a Banach space. To obtain a Banach space you must specify a complete norm. It is a theorem that in a finite dimensional vector space, any norm will be complete. So, to obtain a Banach space structure on $\mathbb R^n$ all you have to do is choose a norm. There are plenty of norms, in fact infinitely many, to choose from. I won't list any here but just mention that in fact once you have a norm you can look at the unit ball in that norm. You'll get a non-empty, bounded, closed, convex, and star-like subset of $\mathbb R^n$. It is not hard to show that given any such subset there is a unique norm giving it as its unit ball. So, you can tailor a norm nearly as freely as you'd like.
In order to make the vector space $\Bbb{R}^n$ (over $\Bbb{R}$) into a Banach space, you'll need to equip $\Bbb{R}^n$ with a norm. That is, a map $$\| \cdot \| : \Bbb{R}^n \to [0, \infty)$$ such that, for all $x, y \in \Bbb{R}^n$, and $\lambda \in \Bbb{R}$,
Typically one equips the Euclidean norm: $$\|(x_1, \ldots, x_n)\| = \sqrt{x_1^2 + \ldots + x_n^2},$$ but there is an enormous variety of possible norms to equip to $\Bbb{R}^n$.
Banach spaces need to be complete, in the sense that Cauchy sequences must converge. Fortunately, every norm on $\Bbb{R}^n$ will make the space complete. (This is not true for infinite-dimensional spaces!)