Bloom filter variant
Let's analyze how many hash bits you need in your new scheme versus a Bloom filter.

First of all, we need to agree about terminology.  I will use $q$ to represent the probability of a false positive.

For a Bloom filter the design problem of choosing $m$ and $k$ given that you want to hold $n$ elements with false positive rate $q$ is solved by $k = -\lg_2 q$ as the number of hashes, and $m=\frac{n k}{\ln 2}$ bits for the table.  (See the wikipedia page or several of the previous cs.se questions about Bloom filters.)  (This works out to a table with maximum entropy: your table is about half 1s and half 0s after $n$ insertions.)  The number of hash bits you need to produce per insertion is $k \lg_2 m$

Now your structure:  Let's assume that the desired $n$ and $q$ are the same.  $k$ and $m$ are going to be about the same, but now $k$ is going to be the average number of bits that remain set after ANDing your $b$ independent $m$-bit hashes.  How big does $b$ need to be?  Well you want the probability of any particular bit in your final hash to be about $k/m = p = 2^{-b}$.  So $b = -\lg_2 k/m = \lg_2 m - \lg_2 k$.  But that means that you have to generate $b m = m(\lg_2m-\lg_2k)$ hash bits, which is way more than $k \lg_2 m$.  For example, if $n = 709$ and $q = 1/16$ then $k = 4$ and $m = 4096$.  So a bloom filter would require $4 \times 12 = 48$ bits of hash, while your scheme requires $b = 10$ and a total of $40960$ bits of hashing.

Worse, there are some elements whose hash works out to 0 (i.e., whenever you look them up the probability of getting a false positive is 100%).  Suppose I AND together 10 different bit vectors of length $m$.  The probability that each bit will be set to 1 is 1/1024.  The probability that all the bits will be set to 0 is $(\frac{1023}{1024})^{4096}\approx e^{-k} \approx 0.0183$.