To find Sylow $p$-subgroups in a group in which you can generally find $p$-elements and normalizers the following algorithm is sometimes used:
Assume $P_i$ is a $p$-subgroup of $G$ (we can take $P_0 = \{1_G\}$ to be the trivial subgroup). If $P_i$ is a Sylow $p$-subgroup, then yay, return $P=P_i$. Otherwise we know by theory* that $P_i$ is not a Sylow $p$-subgroup of $N_G(P_i)$ either. Let $x_i$ be any $p$-element of $N_G(P_i)$ not contained in $P_i$, and set $P_{i+1} = \langle x_i, P_i \rangle$. This is still a $p$-subgroup, since* $x_i$ normalizes $P_i$. Hence we can repeat.
Theory*: If $P_i$ is a not a Sylow $p$-subgroup, then it is properly contained in an as-yet-unknown Sylow $p$-subgroup $P$. Since $P$ is nilpotent, it satisfies the normalizer condition and so $P_1$ is a proper subgroup of the $p$-subgroup $N_P(P_1) \leq N_G(P_1)$. Hence $P_1$ is not a Sylow $p$-subgroup of $N_G(P)$.
Since*: If $H,K \leq G$ and $H$ normalizes $K$, then $HK=KH$ is a subgroup and $|HK| \cdot |H \cap K| = |H| |K|$. In our case these numbers are all powers of $p$.
Application to the symmetric group: You already have $$P_p = \left\langle (1,2,3,\dots,p), (p+1,p+2,p+3,\dots,2p),\dots,(p^2-p+1,p^2-p+2,\dots,p^2-1,p^2) \right\rangle$$
so you just need to find some element that normalizes this group. Hence you need some order $p$ automorphism of the group, say one that cycles the basis elements $$x_i=(p(i-1)+1,p(i-2),\dots,pi)$$ amongst themselves, cyclically, in a cycle of length $p$. Such an element is exactly $$x_p = (1,p+1,2p+1,\dots,p^2-p+1)(2,p+2,2p+2,\dots,p^2-p+2)\cdots(p,2p,3p,\dots,p^2)$$
Generally speaking normalizers can be hard to find, so this algorithm is only used when the more advanced algorithms are not available (which unfortunately is still pretty often AFAIK). However, it just works fine here.
Best Answer
No. Consider the group $S_4$. It is generated by the six transpositions $(12)$, $(13)$, $(14)$, $(23)$, $(24)$ and $(34)$. The order of the group is $4!$ - much less that $2^6=64$
I was more than a bit unhappy with the above example. Its success exploits a kind of sloppiness in the formulation of the question. Namely the fact that I used a non-minimal generating set. To wit, it suffices to use only the 2-cycles $(12)$, $(13)$ and $(14)$ to get all of $S_4$, and the OPs guess is valid for that set of generators. Therefore the above counterexample is lacking in intellectual honesty.
My suggested remedy: