This answer is a bit late, but the problem has quite an elegant solution based on the Fourier transform and I wanted to add it.
The action of the BSC can be modeled as a\pmod 2 sum of the data bit, X, and a random noise bit, N, who's probability distribution is P(N=0)=1-p and P(N=1)=p,
Y = X \oplus N.
The distribution over the output, Y, is a convolution of the quantities P(X) and P(N),
P(Y=y) = \sum_{x \in \{0,1\}} P(X=x)P(N=x\oplus y).
This is a linear transformation of the probabilities. The circulant correlation matrix obtained from one of the vectors is diagonalized by the 2x2 DFT Hadamard matrix,
H = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix},
and the convolution becomes a product of spectra in the transform domain. The sum of L identically distributed noise bits corresponds to exponentiation in the transform domain followed by an inverse transform,
H^{-1}\left(H\begin{bmatrix}1-p\\p\end{bmatrix}\right)^L = \frac{1}{2}\begin{bmatrix}{1 + (1 - 2p)^L}\\{1 - (1 - 2p)^L}\end{bmatrix},
where the second element of the result vector represents the probability P(N_1 \oplus N_2 \oplus N_3 \oplus \ldots \oplus N_L = 1), which is the flip probability for the cascaded BSC.