V.G. Drinfeld: On quasi-Hopf algebras and on a group that is closely connected with $Gal(\overline{\mathbb{Q}}/\mathbb{Q}$), Algebra i Analiz, 2:4 (1990), 149–181
I am not saying it is easy to read but it is definitely well-motivated. By the way, the point of view on $\widehat{GT}$ might also be a bit different from what you expect.
In the very same spirit, but less technical, there is an excellent account of Drinfled's theory by Dror Bar-Natan: On Associators and the Grothendieck-Teichmuller Group I, Selecta Mathematica NS 4 (1998) 183-212 ( http://arxiv.org/abs/q-alg/9606021).
Let me now explain broadly how this works.
Consider the collection $\mathcal B_*=(\mathcal B_n)_n$ consisting of groupoids of parenthesised braids. It is an operad in groupoids (by using the so-called cabling operations).
A more algebroic way to look at it is to consider the free braided monoidal category with one generator $\bullet$, and consider the groupoid of isomorphisms in this category. Objects in this category are actually parenthesisations of $\bullet^{\otimes n}$... so if we fix $n$ we recover $\mathcal B_n$.
Now we would like to define the group $GT$ as the automorphism group of this operad in groupoids $\mathcal B_*$... the main problem is that it is rather small.
The main point is then, for any field $k$, to consider $k$-prounipotent completions $\mathcal B_n(k)$ of $\mathcal B_n$. Now $\widehat{GT}(k)$ is the automorphism group of the operad in $k$-groupoids $\mathcal B_*(k)$.
Using the version of Mac-Lane coherence Theorem for braided monoidal categories, one can prove that such automorphisms are determined by their action on $\mathcal B_3(k)$, and that the relations we have to impose are in $\mathcal B_3(k)$ and $\mathcal{B}_4(k)$.
The relation in $\mathcal B_3(k)$ is the hexagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_2(k)$ to $\mathcal B_3(k)$. The relation in $\mathcal B_4(k)$ is the pentagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_3(k)$ to $\mathcal B_4(k)$.
