Let $a<1<b$. I prove that for large $n$:
1) the probability that a random sequence of length $n$ has at least $B:=b\log_2 n$ consecutive zeroes is at most $n^{1-b}$;
2) the probability that a random sequence of length $n$ does not contain $A:=\lfloor a\log_2 n\rfloor$ consecutive zeroes is at most $e^{-n^{1-a+o(1)}}=O(n^{-M})$ for any $M>0$.
Proofs. 1) For each possible place of $\lceil B\rceil $ consecutive positions consider the event: all positions are 0. Denote these events $E_1,E_2,\dots$. The sum of their probabilities does not exceed
$$\frac{n}{2^B}\leqslant n^{1-b}.
$$
2) Choose $m:=\lfloor n/A\rfloor$ disjoint segments of $A$ consecutive places. The probability that none of them contains only zeroes equals
$$
(1-2^{-A})^m\leqslant e^{-m\cdot 2^{-A}}=e^{-n^{1-a+o(1)}}
$$