Defining $\mathbb{C}$ without defining $\mathbb{R}$ To define $\mathbb{R}$, one approach is to start with $\mathbb{N}$ and then systematically introduce $\mathbb{Z},$ $\mathbb{Q}$, and then $\mathbb{R}$. Alternatively, we define $\mathbb{R}$ using its axiomatic definition that it is a complete ordered field. And then, we can construct $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ within $\mathbb{R}$.
Now, $\mathbb{C}$ is introduced in the first way, by making $\mathbb{R}^2$ as a field. Can we follow the latter approach of defining $\mathbb{C}$ axiomatically, and then construct $\mathbb{R}$ inside it?
 A: You may find this helpful.
However, I disagree with some of the philosophy embedded in your question. You write:

Alternatively, we define R using its axiomatic definition that it is a complete ordered field. And then, we can construct N, Z and Q within R.

With a small caveat, this isn't really correct, insofar as you can't really "construct things axiomatically" - that's not how math works. In particular, to define a thing, you should either:


*

*Construct it out of things that have already known to exist, or

*Characterize it by some constraints and then prove there is a unique thing satisfying those constraints.
Notice the boldface portion above. Suppose you want define $\mathbb{R}$ axiomatically - fine, but you need to prove that there's a unique thing up to isomorphism satisfying your definition. Uniqueness should be easy enough. But how would you prove existence? Well, you'd probably begin by constructing $\mathbb{N}$, and then building your way up to $\mathbb{Q}$ and taking a completion. So this isn't really an alternative. Of course, you could do things differently. For example, you could start of by constructing the surreal numbers recursively, and then obtain $\mathbb{R}$ as a subset of those - that would be a different way of constructing $\mathbb{R}$. But at the end of the day, you still haven't "constructed $\mathbb{R}$ axiomatically" - you've just constructed it in a different way.
The caveat I mentioned earlier is that you can replace the axiom of infinity in ZFC with an axiom that instead posits the existence of a real numbers object, thereby obtaining a new system, let's call it ZFCR. It seems likely that ZFCR is equivalent to ZFC, for the reasons you give.
A: Perhaps the unique algebraically and topologically complete partially ordered field of characteristic $0$? 
However, I am not sure it is true - dropping from ordered to partially ordered opens up a lot of freedom that may allow other contenders.
I am modifying this answer (the idea above is highly doubtful) to address Arvind's questions in the July 6 comment below.
Here is an outline of one approach to prove that $\Bbb C$ is, up to isomorphism, the only connected, locally compact topological field of characteristic $0$ which remains connected when $0$ is removed. There may be easier approaches, and I have not gone through the steps myself to make sure there are no significant hurdles that I am overlooking, but this is the approach I would try:
Let $\mathscr C$ be a connected, locally compact topological field of characteristic $0$ for which $\mathscr C \setminus \{0\}$ is also connected. We need to show that $\scr C$ is isomorphic to $\Bbb C$ as topological fields.


*

*Let $\scr N\subset C$ be the smallest set containing $0$ and $1$, and closed under addition. One can inductively define a well ordering on $\scr N$  by the requirement that $n < n + 1$. Using this well-ordering, define an isomorphism of monoids with $\Bbb N$ that also preserves multiplication. (Note that this requires $\scr C$ to be of characteristic $0$ to work.)

*Let $\scr Z$ be the smallest set containing $\scr N$ closed under subtraction. Extend the order on $\scr N$ to $\scr Z$. The order-preserving isomophism of monoids between $\scr N$ and $\Bbb N$ extends to an order-preserving isomorphism of unital rings between $\scr Z$ and $\Bbb Z$.

*Let $\scr Q$ be the smallest field in $\scr C$. Note that it must contain $\scr Z$. Show that the order and isomorphism also extend to $\scr Q$, where the isomorphism is a isomorphism of topological fields.

*Use the continuity of the $+, \cdot$ operators to show that the topology on $\scr Q$ induced by the order is the same as the subspace topology from $\scr C$

*Let $\scr R$ be the closures of $\scr Q$. Show that it is also a field, and the order and isomorphism extend as well. And again that the order topology is the same as the subspace topology. Note that this also shows $\scr R$ is complete, since $\Bbb R$ is.

*Note that $\scr C$ is a vector space over the subfield $\scr R$. Use local compactness at $0$ to prove that the vector space must be finite dimensional.

*Since $\scr R$ is isomorphic to $\Bbb R$, we know that $\scr R[x]/(x^2 + 1)$ is algebraicly closed. Hence the only finite dimensional extensions of $\scr R$ have dimension $1$ or $2$. If $\scr C$ has dimension $1$ over $\scr R$, it must be $\scr R$ itself.

*Note that $\scr R \setminus \{0\}$ is disconnected (since it is isomorphic to $\Bbb R$), but $\scr C \setminus \{0\}$ is connected. Therefore $\scr C \ne \scr R$.

*Conclude that $\scr C$ is a two-dimensional extension of $\scr R$, and therefore isomorphic to $\scr R[x]/(x^2 + 1)$, which contains two elements $i$ satisfying $i^2 = -1$. Choose one of them arbitrarily to be "$i$", and use it to extend the isomophism between $\scr R$ and $\Bbb R$ to an isomorphism between $\scr C$ and $\Bbb C$.

