To answer the first question in the OP, note that the magnitude of the vector $\vec v=\frac{\hat r}{r^2}$ is given by
$$|\vec v|=\frac{1}{r^2}$$
Inasmuch as $|\vec v|$ decays with increasing $r$, the field lines decay commensurately.
To answer the second question, note that for any closed and smooth surface $S$ that encompasses the origin, the outward flux $\Psi$ is
$$\Psi =\oint_S \vec v\cdot \hat n\,dS=4\pi$$
This is surprising since $\nabla \cdot \vec v=0$ for all $r\ne 0$. Therefore, in terms of both Lebesgue and Riemann integrals, $\int_V \nabla \cdot \vec v\,dV=0$. But naïve application of the divergence theorem would show that
$$0=\int_V \nabla \cdot \vec v\,dV=\oint_S \vec v\cdot \hat n\,dS=4\pi \tag 1$$
The reason that the logic used in $(1)$ is flawed is that the divergence theorem is inapplicable since $\nabla \cdot \vec v$ does not exist at the origin.
NOTE: Broadening the Interpretation of $\nabla \cdot \left(\frac{\hat r}{r^2}\right)$
In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as
$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)\sim 4\pi \delta(\vec r)} \tag 2$$
where
$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 3$$
Note that in $(3)$, $\vec \psi(\vec r;a=0)=\frac{\hat r}{r^2}$ represents the electric field from a point charge $q=4\pi \epsilon_0$.
Hence, we interpret $(2)$ as a representation of $\nabla \cdot \left(\frac{\hat r}{r^2}\right)$ in the sense of distributions. The symbol $\sim$ in $(2)$ means that for any suitable test function $\phi(\vec r)$, we have
$$\lim_{a\to 0}\int_{V} \phi(\vec r)\,\nabla \cdot \vec \psi(\vec r;a)\,dV=4\pi \phi(0)$$
whenever the region $V$ contains the origin $\vec r=0$.
