It is apparently assumed that the $Z_{ij}$'s are independent, as we will do here -- since otherwise hardly anything can be said. Suppose also that $m\ge2$ and $0<p<1$.
The $ab$-entry of the matrix $Y:=XX^\top$ is \begin{equation} Y_{ab}=\sum_{r\in[m]}X_{ar}X_{br} =\sum_{r\in[m]}\frac{Z_{ar}Z_{br}}{\sum_{k\in[m]}Z_{ak}\sum_{k\in[m]}Z_{bl}}, \end{equation} where $[m]:=\{1,\dots,m\}$ and $a,b$ are in $[n]$; here we assume the convention $\frac00=0$. So, \begin{equation} EY_{ab}=m \,E\frac{Z_{a1}Z_{b1}}{\sum_{k\in[m]}Z_{ak}\sum_{k\in[m]}Z_{bl}}. \end{equation} If $a\ne b$ then \begin{equation} \begin{aligned} EY_{ab}&=m\,E\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}} \, E\frac{Z_{b1}}{\sum_{k\in[m]}Z_{bl}} \\ &=m \Big(E\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}}\Big)^2 \\ &=m p^2\Big(E\frac1{1+\sum_{k\in[m]\setminus\{1\}}Z_{ak}}\Big)^2 \\ &=m p^2\Big(E\frac1{1+B_{m-1,p}}\Big)^2 \\ &=\frac{\left(1-q^m\right)^2}{m}, \end{aligned} \end{equation} where $q:=1-p$ and $B_{m-1,p}$ is a binomial random variable with parameters $m-1,p$. Similarly, the diagonal entries of the matrix $EY$ are \begin{equation} \begin{aligned} EY_{aa}&=m E\Big(\frac{Z_{a1}}{\sum_{k\in[m]}Z_{ak}}\Big)^2 \\ &=m p\,E\Big(\frac1{1+\sum_{k\in[m]\setminus\{1\}}Z_{ak}}\Big)^2 \\ &=m p\,E\Big(\frac1{1+B_{m-1,p}}\Big)^2 \\ &=mp q^{m-1} \, _3F_2\left(1,1,1-m;2,2;-\frac{p}{q}\right), \end{aligned} \end{equation} where $_3F_2$ is the hypergeometric function.