Absolute summability will allow you to show that the sequence (in $n$)
$$
X_t^n = \sum_{j=-n}^{n} \psi_jZ_{t-j}
$$
has a mean-square limit (we are not talking about almost-sure convergence, here). That is, we want to show that there exists some $X_t$ (we don't know it exists yet,because it's an infinite sum) such that
$$
\mathbb{E}[|X_t^n -X_t|^2] \to 0
$$
as $n \to \infty$.
Often it is easier to verify that the sequence $X_t^n$ is Cauchy, which is an equivalent condition. This means that $\mathbb{E}[|X_t^n -X_t^m|^2] \to 0$ as $m,n \to \infty$.
Here's the proof. For $n > m > 0$
\begin{align*}
&\mathbb{E}[|X_t^n -X_t^m|^2] \\
&= \mathbb{E}\left[\left| \sum_{m < |j| \le n} \psi_jZ_{t-j}\right|^2\right] \\
&= \sum_{m < |i| \le n} \sum_{m < |k| \le n} \psi_i \psi_k\mathbb{E}[Z_{t-i }Z_{t-k}] \\
&\le\sum_{m < |i| \le n} \sum_{m < |k| \le n} |\psi_i| |\psi_k|
|\mathbb{E}[Z_{t-i}Z_{t-k}]| \tag{triangle ineq.}\\
&\le\sum_{m < |i| \le n} \sum_{m < |k| \le n} |\psi_i| |\psi_k|
(\mathbb{E}[Z_{t-i}^2])^{1/2} (\mathbb{E}[Z_{t-k}^2])^{1/2} \tag{Cauchy-Schwarz}\\
&= \text{Var}(Z_t) \left( \sum_{m < |j| \le n} |\psi_j| \right)^2 \tag{stationarity of $Z_t$}\\
&\to 0 \tag{absolute summability}.
\end{align*}
Only after you know that $X_t$ exists, can you show the order of taking the limit and expectation doesn't matter. Or in other words, you can show that
- $E[X_t^n] \to EX_t$,
- $E[|X_t^n|^2] \to E[X_t^2]$, and
- $E[X_t^nX_s^n] \to E[X_tX_s]$;
but existence comes first.
Edit:
To prove as convergence, we can use the Borel-Cantelli lemma. Pick $\epsilon > 0$ and call
$$
A_n = \{|X_t^n - X_t| > \epsilon \} = \left\{ \left|\sum_{|j|>n} \psi_j Z_{t-j}\right| > \epsilon\right\}.
$$
Using the same reasoning above
\begin{align*}
\sum_{n=1}^{\infty} P(A_n) &\le \epsilon^{-2}\sum_{n=1}^{\infty} E\left[\left|\sum_{|j|>n} \psi_j Z_{t-j}\right|^2\right] \\
&\le \text{Var}(Z_t) \left( \sum_{j \in \mathbb{Z}} |\psi_j| \right)^2\\
&< \infty.
\end{align*}
So $X_t^n \overset{as}{\to} X_t$ for each $t$.