This is a very interesting question and I guess that a general answer is unknown, already in the case $i=1$. Let me just make the following

Remark. There exists no upper bound on $\textrm{rank } NS(X)$ which is independent on the dimension.

In fact, let us consider a complex Abelian variety $X$ of dimension $g$ such that $End_{\mathbb{Q}}(X)$ is a totally real number field of degree $g$ over $\mathbb{Q}$. These varieties do exist and the general one is simple, see [Birkenhake - Lange, Chapters 5 and 9].

Therefore it is known that

$\rho(X) =\textrm{rank } \textrm{NS}(X) = \textrm{rank } End^s_{\mathbb{Q}}(X)=g$,

where $End^s_{\mathbb{Q}}(X)$ denotes the subgroups of elements which are symmetric with respect to the Rosati involution.

On the other hand, in a simple Abelian variety any effective divisor is ample, so two effective divisors always intersect and $X$ satisfies Property 1.

This is a very interesting question and I guess that a general answer is unknown, already in the case $i=1$. Let me just make the following

Remark. There exists no upper bound on $\textrm{rank } NS(X)$ which is independent on the dimension.

In fact, let us consider a complex Abelian variety $X$ of dimension $g$ such that $End_{\mathbb{Q}}(X)$ is a totally real number field of degree $g$ over $\mathbb{Q}$. These varieties do exist and the general one is simple, see [Birkenhake - Lange, Chapters 5 and 9].

Therefore it is known that

$\rho(X) =\textrm{rank } \textrm{NS}(X) = \textrm{rank } End^s_{\mathbb{Q}}(X)=g$,

where $End^s_{\mathbb{Q}}(X)$ denotes the subgroups of elements in $End_{\mathbb{Q}}(X)$ which are symmetric with respect to the Rosati involution.

On the other hand, in a simple Abelian variety any effective divisor is ample, so two effective divisors always intersect and $X$ satisfies Property 1.