If we have $m$ subgroups of order $p^n$ at least how many different elements we can have. For example in case of $n=1$ we have $m(p-1)$ because a subgroup of order $p$ is cyclic so different subgroups share just the identity. In case of $n=2$ am not sure I think at least $m(p-1)$ right?
Am studying slow theorem and sometimes we need to figure how many different elements we could have from $n_p$ ${\rm Sly}_P$ subgroups
In full generality, very little can be said about this besides the obvious. There are examples where the common intersection is as big as possible (index $p$) and examples where the pairwise intersection is as small as possible (trivial).
For example, take $p=2$, $m$ odd and $G=C_2^{n-1}\times D_{2m}$ (where $C_2$ is the cyclic group of order $2$ and $D_{2m}$ is the dihedral group of order $2m$). Then Sylow $2$-subgroups of $G$ have order $2^n$, there are $m$ of them, and they all intersect in a common subgroup of index $2$, so as big as possible.
At the other end of the spectrum take for example, $A_5$ has $5$ Sylow $2$-subgroups, of order $4$, and the pairwise intersections are trivial.