Okay, as it turns out the original post had the right idea, but after a crazy day my execution was a little off.
As I understand the problem, (B_1) is a bit of a red herring. We're looking essentially for the probability that the second Brownian increment (x=B_2-B_1 > 0) and then the third Brownian increment (y = B_3-B_2 < -x). Given (x), the probability (P(y < -x) = N(-x)); all we have to do is integrate over all (x) that satisfy the initial condition, weighted by the probability of that outcome, to get the answer.
In other words, we want to evaluate the integral:
(\int_{0}^{+\infty} N'(x)N(-x) dx)
By symmetry, we see that...
(\int_{0}^{+\infty} N'(x)N(-x) dx = \int_{0}^{+\infty} N'(x)[1-N(x)] dx)
(=\int_{0}^{+\infty}N'(x)dx - \int_{0}^{+\infty} N'(x)N(x)dx = \frac{1}{2} - \int_{0}^{+\infty} N'(x)N(x)dx)
That second integral can be done pretty easily (try it by parts) and is equal to (\frac{3}{8}), giving the eventual answer of (\frac{1}{8}).
That final answer actually makes intuitive sense. After all, the simple probability that (x > 0) and then (y < 0) is (\frac{1}{4}). Now, take some path (i) satisfying this condition, and suppose arbitrarily that (x_i > -y_i); then in this case our condition is not satisfied. However, for each such path we can identify an equally probable path by just swapping the values (x_{i'} = -y_i, y_{i'} = -x_i) that does satisfy the condition.
So, among our universe of (\frac{1}{4}) of all possible paths, there is a one-to-one correspondence between paths that satisfy the desired conditions and ones that don't. Since the set of paths ending exactly at zero is of zero measure, we conclude that the probability of satisfying the condition we want is (\frac{1}{8}).