Wrapping up the comments under the original question: One viewpoint of (differentiable) manifolds is that manifolds are those objects that can be approximated locally by a linear space, the "model space", so that concepts of calculus can be transported from the model space to the manifold.
The model space needs to be at least a topological vector space, a vector space that is also a topological space such that the algebraic operations are continuous, to make sense of notions of calculus. You first need to be able to define e.g. the derivative of a path in your model space before you can define the derivative of a path in the manifold.
If you take a curve in your model space $E$:
$ c: \mathbb{R} \to E $
you'd need to be able to write down the definition of the derivative c':
c'(t) := \lim_{s \to 0} \frac{1}{s} (c(t+s) - c(t))
In order to make sense of this formula you need the linear structure, a topology, and the continuity of addition and scalar multiplication.
The simplest example of a topological vector space is of course $\mathbb{R}^n$. It is a theorem that on a finite dimensional vector space there is one and only one topology that turns the vector space into a topological vector space, which in the case of $\mathbb{R}^n$ is the canoncial topology induced by, for example, the Euclidean norm.
Therefore, when one talks about differentiable manifolds modelled on $\mathbb{R}^n$, $\mathbb{R}^n$ is always equipped with this unique topology. The same applies to complex manifolds modelled on $\mathbb{C}^n$, too, of course. Things get more interesting for model spaces that are infinite dimensional, like Fréchet spaces. But as long as you are talking about differentiable manifolds modelled on finite dimensional spaces, the topology is fixed by the requirement that the model space needs to be a topological vector space.