I intuitively got that the fundamental group of the topological space given by considering $4$ spheres pairwise tangent in $ \mathbb{R}^3$ is $\langle a,b,c\rangle$.
My approach was that if you consider the fundamental group of $3$ spheres pairwise tangent you get the same one as the circumference. Applying this to the $4$ triplets of spheres you get when considering this space we intuitively get 4 pairwise tangent circles, that can be deformed to a $1$-skeleton of a tetrahedron, that I know to have fundamental group equal to $\langle a,b,c \rangle$. The problem is, $3$ tangent spheres are homotopy equivalent to the circumference, not homeomorphic and I don't know if the union of subspaces has the same homotopy type as the union of homotopy-equivalent spaces. Another approach I have tried is trying to find an equivalence between this space and a sphere without $4$ points, but I'm having a hard time finding such maps.
Topology – Fundamental Group of Four Pairwise Tangent Spheres
algebraic-topologygeneral-topologylow-dimensional-topology
Related Solutions
It is a common misconception that there is no nice combinatorial description of the earring space (or other wild 1-dimensional spaces like the Sierpinski carpet or Menger curve for that matter). Once you get used to the construction, it is actually not too bad at all. I would argue that it is far more tractable than the homotopy groups of spheres. Moreover, this group has become quite important in infinite group theory since in many ways it is the non-abelian Specker group.
Pece is right that the way to deal with $\pi_1(H)$ is as a subgroup of an inverse limit of free groups. This representation of $\pi_1(H)$ is secretly use "shape theory." I give a friendly introduction to the earring group in the blog post:
http://wildtopology.wordpress.com/2013/11/23/the-hawaiian-earring/
In more recent blog posts I have walked step by step through De Smit's subtle argument that $\pi_1(H)$ is not free (which Pece references).
Not a complete answer
Hatcher shows two unlinked circles have a fundamental group $\mathbb{F}_2$ which is humungous - it's the free group of all possible wors in two letters. The fundamental group of two linked circles is simply $\mathbb{Z} \oplus \mathbb{Z}$ - we can exploit the linking of the rings in order to express any entanglement as a special type.
In order to find the homotopy group of the Borromean rings you may need to draw as many as 4 rings.
3 pairwise linked circles will have three generators, looping ones around each of the possible three circles. Do these generators commute? Sure we can imagine one the circles disappearing and getting two circles. We have shown:
$$ \pi_1(3\text{ linked circles}) \subseteq \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z}$$
Best Answer
Consider the space $X$ to be the 1-skeleton of a tetrahedron, with a sphere glued to every vertex of the skeleton.
It is possible to write down a homotopy equivalence between your space and $X$. By Seifert-Van Kampen you can then show that the fundamental group of $X$ must be $\langle a,b,c\rangle$ as you said.