It depends what you mean by classification. Let's start with the example of the free group of rank $n$, $F_n$. The normal subgroups of $F_n$ correspond to $n$-generator groups with marked generating sets, which of course is a hopelessly complicated set. So in some sense, the normal subgroups of $F_n$ are not classifiable. On the other hand, we have:

**Greenberg's Theorem:** If $N\lhd F_n$ then precisely one of the following holds.

- $N$ is trivial.
- $N$ is of finite index in $F_n$.
- $N$ is infinitely generated.

Of course, the second option is pretty complicated, and the third even more so. But you might call this a sort of 'classification'.

Now, the pure braid group $PB_n$ admits a natural map onto $PB_{n-1}$, by forgetting a strand. As observed above, $B_3$ surjects $\mathbb{Z}/2*\mathbb{Z}/3$, which has a free subgroup of finite index. So every $B_n$ virtually surjects a free group. This indicates that the normal subgroups of $B_n$ are almost as complicated as the normal subgroups of $F_m$.

The OP has now indicated that he is interested in the normal subgroups *of finite index*. The answer above shows that we at least have the following:

Let $n\geq 3$. For every finite group $Q$ there is a subgroup $N\lhd B_n$ of finite index with $Q\hookrightarrow B_n/N$.

That said, there are probably many more normal subgroups of finite index. Here's a question in this direction to which I don't know the answer.

Are braid groups residually finite simple?

Further edit:

I think this is what you're looking for. Here's an article by McReynolds, giving a proof that pure braid groups have the congruence subgroup property. This can be thought of as a classification of the subgroups of finite index.