Not quite, the cost of RSA encryption is O(k^3), where k is the total number of bits, not the number of 1's. But it is cheaper to have less 1 bits, as (abusing big-O notation), the cost is more like O(3k^3-z^3), where z is the number of 0 bits. That's why some people are using 2^32+1 instead of 2^32-1.
In theory the size of e doesn't matter, but a larger e does make it harder to brute force the message text.
You're mixing up the number of bits in the RSA modulus n, and the number of bits in the exponent e.
Integer modular multiplication is O(n^2)[1]. Binary exponentiation, the most common of algorithms, requires O(e) modular multiplications. It is therefore trivial to get to the O(n^3) figure of general modular exponentiation.
RSA encryption with e = 3 is much simpler, consisting of exactly 2 modular multiplications. That makes it O(n^2).
Note that an exponent with very low hamming weight can still be O(n^3). For example e = 2^(n-1) + 1 requires n multiplications, therefore O(n^3).
[1] If you want to go all theoretical, it can be as low as O(n log n log log n) using Schoenhage-Strassen. But that is not practical for 1024--4096 bit integers.
In theory the size of e doesn't matter, but a larger e does make it harder to brute force the message text.