Is every torsion-free abelian group isomorphic to $\mathbb{Z}^n$? Is every torsion-free abelian group isomorphic to $\mathbb{Z}^n$ for some natural $n$?
I think it is true just for finitely generated abelian torsion-free groups...
 A: Every finitely-generated torsionfree abelian group is isomorphic to $\mathbb Z^n$ for some $n$, but the proof crucially relies on the finiteness hypothesis (as most of them use induction, one way or another).
In fact, there are torsionfree abelian groups that are not isomorphic to $\mathbb Z^{(I)}$ for any $I$, or to $\mathbb Z^I$ for any $I$ (the two possible generalizations one may think of for infinite $I$). 
As stated in the comments, $\mathbb Q$ is one such example, but so is the group of $p$-adic integers $\mathbb Z_p$ or the additive group of the ring $\mathbb Z[1/p]$. 
In $\mathbb Z^{(I)}$ or $\mathbb Z^I$, there are no elements that are divisible by arbitrarily high integers, which is what fails in these examples. 
(If you know homological algebra, here's some food for thought : over $\mathbb Z$, torsion-free modules are precisely the flat modules; and modules of the form $\mathbb Z^{(I)}$ are precisely the projective modules. In these terms, your question amounts to "over $\mathbb Z$, does flat imply projective ?")
A: You have all the elements to try $\mathbb{Z}^\mathbb{N}$ with componentwise addition.
A: To further clarify the question and the mentioned comments and in order to provide a complete answer:
An elementary argument shows that the following proposition holds:  
Proposition. Every finitely-generated torsion-free abelian group is isomorphic to $\mathbb{Z}^{n}$ for some $n\geq 1.$
proof. Suppose that $ G $ is generated by $ \{a_{1},a_{2},\ldots,a_{n}\}; $ i.e., we have:
$$G:=\Big\{\sum_{i=1}^{n}k_{i}a_{i}: k_{i}\in\mathbb{Z}\Big\}.$$
Let $ \bar{\mathbf{1}}_{i} $ be the element in $ \mathbb{Z}^{n} $ all of whose components are equal to zero except for the $ i $th component that equals $ 1. $ It is easy to verify that the function that maps each $ a_{i} $ to $ \bar{\mathbf{1}}_{i} $ is a group isomorphism.$\qquad\blacksquare$
And, as mentioned in the comments, the additive group of the rationals is an example of a torsion-free abelian group that is not finitely-generated. 
Looking at $ \langle\mathbb{Q},+,0\rangle$ from another point of view, it is an ordered divisible group. It can be shown that an abelian group is torsion-free if and only if it is orderable (some hints are available in the link below about an argument that uses the proposition above). Finally, an interesting question might be the following:  
Question. Is there an abelian divisible torsioned group?
More detailed explanations can be found in the following question:
Is there a natural example of a divisible torsioned (= periodic) abelian group?
