When is asymptotic stability preserved in this sequence of dynamical systems?  Setup:  Suppose that for each $a>0$, $(a,0)$ is an asymptotically stable steady state of the autonomous system
$$
\begin{cases}
\frac{d}{dt}{z}_1(t) = f_1\big(z_1(t),z_2(t);a\big)\\
\frac{d}{dt}{z}_2(t) = f_2\big(z_1(t),z_2(t);a\big)\\
\end{cases} \qquad (*)
$$
where $f_i:\mathbb{R}^2\times$  $\mathbb{R}_{>0}$ $\to\mathbb{R}$ is continuously differentiable $(i=1,2)$.

 Remark:  note that this implies that -- given system $(*)$ -- the  basin of attraction of $(a,0)$
$$ \left\{\vec{z}_0\in\mathbb{R}^2: \big(z_1(0),z_2(0)\big)\,\texttt{=}\,\vec{z}_0 \Rightarrow \lim_{t\to\infty}z_1(t)=a \text{ and } \lim_{t\to\infty}z_2(t)=0 \right\} $$
has strictly positive Lebesgue measure $\forall a>0$ (details ).

Assume that $\lim\limits_{a\to\infty}f_{i}(\cdot;a)=\hat{f}_{i}(\cdot)\in\mathcal{C}$ pointwise ($i=1,2$),
and define the autonomous system
$$
\begin{cases}
\frac{d}{dt}{z}_1(t) = \hat{f}_{1}\big(z_1(t),z_2(t)\big)\\
\frac{d}{dt}{z}_2(t) = \hat{f}_{2}\big(z_1(t),z_2(t)\big)\\
\end{cases} \qquad (**)
$$
 Question:  Are any additional assumptions needed for the following statement to be true?

$``$Assuming $(z_1(t),z_2(t))$ obey system $(**)$, the set
$$ \left\{\vec{z}_0\in\mathbb{R}^2: \big(z_1(0),z_2(0)\big)\,\texttt{=}\,\vec{z}_0 \Rightarrow \lim_{t\to\infty}z_1(t)=\infty \text{ and } \lim_{t\to\infty}z_2(t)=0 \right\} $$ has strictly positive Lebesgue measure.$"$

If so, which assumptions are sufficient (and necessary) for this to be true?"
 A: The following example answers to your first question: "Are any additional assumptions needed for the following statement to be true?" Yes.
Example. Define the vector-field as $$f_1\left(z_1,z_2;a\right)\overset{\Delta}=\left(\frac{1}{a}z_1\right)\left(a-z_1-z_2\right),$$
$$f_2\left(z_1,z_2;a\right)\overset{\Delta}=\left(\frac{1}{2a}z_2\right)\left(a-z_1-z_2\right)-\gamma z_2,$$
for some $\gamma>0$. Define the set $\mathcal{I}_a:=\left\{(x,y)\in \mathbb{R}^2_{\geq 0}\,:\,x+y\leq a\right\}$. This set is invariant w.r.t. the dynamics defined by the vector field $F=(f_1,f_2)$. This invariance is fairly straightforward to check. Further, remark that ${\sf Leb}\left(\mathcal{I}_a\right)=a^2/2$. Define $\phi_a(t;z_1(0),z_2(0))$ to be the flow (or solution) to this ODE with initial condition $(z_1(0),z_2(0))$.
Theorem. $\phi_a(t;z_1(0),z_2(0))\overset{t\rightarrow \infty}\longrightarrow (a,0)$, for any initial condition $(z_1(0),z_2(0))\in\mathcal{I}_a\setminus \left\{(0,0)\right\}$, for any $a>0$.

Proof. Define $w(t):=z_2(t)/z_1(t)$. We have
$$w'(t)=\frac{z'_2(t)z_1(t)-z'_1(t)z_2(t)}{z_1^2(t)}=w(t)\left(-\frac{1}{2a}(a-z_1(t)-z_2(t))-\gamma\right)$$
and from Gronwall's inequality, we have
$$w(t)\leq  w(0)e^{-\frac{1}{2a}\int_0^t (a-z_1(s)-z_2(s))ds-\gamma t} \leq w(0)e^{-\gamma t}\overset{t\rightarrow \infty}\longrightarrow 0,$$
where the latter inequality holds in view of the invariance of the set $\mathcal{I}_a$, i.e., $a-z_1(s)-z_2(s)\geq 0$ for all $s$. Therefore, $z_2(t)\overset{t\rightarrow \infty}\longrightarrow 0$. It is easy to check that $z_1(t)\overset{t\rightarrow \infty}\longrightarrow a$ (i.e., $z_1$ has the same asymptotic behaviour as if $z_2(t)=0$ for all $t$, that is, $z'_1(t)=(\frac{1}{a}z_1(t))(a-z_1(t))$).

However, $f_1(z_1,z_2;a)\overset{a\rightarrow \infty}\longrightarrow z_1$ and $f_2(z_1,z_2;a)\overset{a\rightarrow \infty}\longrightarrow (1/2-\gamma)z_2$. This means that the limiting vector field is given by $\widehat{f}_1(z_1,z_2)=z_1$ and $\widehat{f}_2(z_1,z_2)=(1/2-\gamma)z_2$ and therefore, the flow of the limiting vector-field $\phi_{\infty}(t,z_1(0),z_2(0))\longrightarrow (\infty,\infty)$, if $\gamma<1/2$ or $\phi_{\infty}(t,z_1(0),z_2(0))\longrightarrow (\infty,-\infty)$, if $\gamma>1/2$, regardless of the initial condition (except the origin), i.e., for any $(z_1(0),z_2(0))\in \mathbb{R}^2\setminus \left\{(0,0)\right\}$.
This further goes by saying that in general the limits do not commute (for almost all initial conditions)
$$\lim_{t\rightarrow \infty} \lim_{a\rightarrow \infty} \phi_{a}(t;z_1(0),z_2(0))=(\infty,\infty)\neq (\infty,0)=\lim_{a\rightarrow \infty} \lim_{t\rightarrow \infty} \phi_{a}(t;z_1(0),z_2(0))$$
Summary. This is an example of an ODE where the Lebesgue measure of the Basin of attraction to $(a,0)$ scales to infinity with $a$ (since it is at least $a^2/2$) and thus it is bounded away from zero for all $a$, but at infinity, it collapses to zero, i.e., the Basin of $(\infty,0)$ is a null-set (perhaps an empty set).
Remark. Regarding your second question: "If so, which assumptions are sufficient (and necessary) for this to be true?" I am not sure, but there should be some specific conditions as the counter-example provided refers to a smooth vector field. The vector-field is also monotonous in an appropriate sense. The limiting (in $a$) vector-field is also smooth. In other words, this example entails a lot of regularity. It is, in a sense, a quite pessimistic example for your purpose.
