I would claim that the "right" way to understand the Clifford group abstractly is to realize that "it is a projective representation of $\mathbb{Z}_2^{2n}\rtimes \mathrm{Sp}(\mathbb{Z}_2^{2n})$" as explained below:

- Start with the discrete $2n$-dimensional vector space `V:=$\mathbb{Z}_2^{2n}$.
- Next, we introduce a symplectic inner product on $V$. Here's one way of doing it: take a basis $e_1, \dots, e_{2n}$ of $V$. We divide this basis into two blocks by setting $p_i = e_i$ and $q_i = e_{n+i}$ for $i=1,\dots n$. Now the symplectic inner product is defined by the relations $[p_i, q_j] = [q_j, p_i] = \delta_{i,j}, [p_i, p_j]=0, [q_i, q_j]=0$ and their linear extensions. (A symplectic inner product is anti-symmetric, so we'd expect $[p_i,q_i]=-[q_i,p_i]$. But since we're working modulo 2 where $-1=+1$, there's no manifest negative sign in the definition above).
- Almost there. Now that we have a symplectic geometry on $V$, we can define the symplectic group $\mathrm{Sp}(V)$, i.e. those linear operations $S$ on $V$ which preserve the inner product in that $[Sv, Sw]=[v,w]$ for all vectors $v,w\in V$.
- Both $\mathrm{Sp}(V)$ and $V$ itself act on $V$ (the latter by addition). Let $G$ be the group generated by these two actions. It's a subgroup of the affine group - namely those affine operations, where the linear part is symplectic. I've heard people calling it the "Symplectic-Affine Group". Technically, it's a semi-direct product between $V$ and $\mathrm{Sp}(V)$.

Anyway, the Clifford group is a faithful projective representation of this Symplectic-Affine group $G$. In other words, the Clifford group up to phases is "just" $V\rtimes \mathrm{Sp}(V)$. And, yes, that's exactly the discrete version of the well-known group of phase-space symmetries in continuous-variable systems. And, no, that's no coincidence.

The question as you asked it would boil down to giving a presentation of the relevant discrete symplectic group in terms of generators and relations. I'd be surprised if there were an insightful way of doing that. So I think you should alter your question to read "(a) what's the best abstract way of understanding the Clifford group and (b) does it involve generators and relations?" to which I would answer: "(a) see above and (b) no, it doesn't".

Everything I said has been discovered and re-discovered many times over in different communities, including mathematical physics ("Stone-von Neumann Theorem" and all that), number theory (Weil), engineering ("oscillator group"), and quantum information ("Clifford group"). Since there are too many references to list them, I just cite my own paper: http://arxiv.org/abs/quant-ph/0602001 and references therein.

Some previous answers referred to spin groups. There's also a "Clifford group" in this context which has, however, nothing to do with the Clifford group as defined in quantum information.