Is this question being asked just out of curiosity? As far as I know, the series $\sum_{n \geq 0} n!$ is not important in $p$-adic analysis.

For $j \in \mathbf N$ and $q \not= 1$, let $(j)_q = (q^j-1)/(q-1) = 1 + q + \cdots + q^{j-1}$. As $q \rightarrow 1$ we have $(j)_q \rightarrow j$ so set $(j)_1 = j$ for $j \in \mathbf N$. Set $(n)!_q = (n)_q(n-1)_q\cdots (2)_q(1)_q$ for $n \in \mathbf Z^+$ and $(0)!_q = 1$. Then $(n)!_1 = n!$. You're looking at $\sum_{n \geq 0} (n)!_q$ as $q \rightarrow 1$ in $K$ and you want to show this series tends to $\sum_{n \geq 0} n! = \sum_{n \geq 0} (n)!_1$ as $q$ tends to $1$, of course after we know that $\sum_{n \geq 0} (n)!_q$ converges in $K$ for $q$ close to $1$. It will turn out "close" can mean $|q-1| < 1$.

For $j \in \mathbf N$ and $|q-1| < 1$, show $|(j)_q - j| \leq |q-1|$, so in the ring $\mathcal O_K/(q-1)$, $(j)_q \equiv j$ for all $j \in \mathbf N$ and thus $(n)!_q \equiv n!$ for all $n \in \mathbf N$. Therefore
$$
\left|\sum_{n = 0}^N (n)!_q - \sum_{n = 0}^N n!\right| = \left|\sum_{n = 0}^N \left((n)!_q - n!\right)\right| \leq |q-1|.
$$
Since $K$ is a local field we know $n! \rightarrow 0$ in $K$ as $n \rightarrow \infty$. *If* we knew $(n)!_q \rightarrow 0$ as $n \rightarrow \infty$ too then we could pass to the infinite series and say
$$
\left|\sum_{n \geq 0} (n)!_q - \sum_{n \geq 0} n!\right| = \left|\sum_{n \geq 0} \left((n)!_q - n!\right)\right| \leq |q-1|.
$$
Then let $q \rightarrow 1$ and you'd have what you were asking about. Does $(n)!_q \rightarrow 0$ in $K$ as $n \rightarrow \infty$?

The field $K$ contains some $\mathbf Q_p$. For that prime $p$, if $|q-1| < (1/p)^{1/(p-1)}$ and $q \not= 1$ then check (using $p$-adic exponential and logarithm, or other methods) that $|(j)_q| = |j|$ for all $j \in \mathbf N$; it's obvious at $q=1$. Then $|(n)_q!| = |n!|$ for all $n \in \mathbf N$, so $(n)!_q \rightarrow 0$ as $n \rightarrow \infty$ when $|q-1| < (1/p)^{1/(p-1)}$, which is good enough since such $q$ are a neighborhood of $1$ and you're interested in a limit as $q \rightarrow 1$.

But actually we have $(n)!_q \rightarrow 0$ as $n \rightarrow \infty$ under the weaker condition $|q-1| < 1$. For such $q$ we can't use the $p$-adic exponential and logarithm and we won't have $|(j)_q| = |j|$ all the time, but that's okay. For all $j$ we have $|(j)_q| \leq 1$, so $|(n)!_q| \leq |(p^r)_q|$ where $p^r \leq n < p^{r+1}$. Since $n \rightarrow \infty$ is the same as $r \rightarrow \infty$, it suffices to show $(p^r)_q \rightarrow 0$ in $K$ as $r \rightarrow \infty$ when $|q-1| < 1$. At $q = 1$ as have $(p^r)_q = p^r$, which clearly tends to $0$ in $K$ as $r \rightarrow \infty$, so let $0 < |q-1| < 1$. Then our task is the same as showing $q^{p^r} \rightarrow 1$ in $K$ as $r \rightarrow \infty$, and that is a special case of the continuity of the function $q^x = \sum_{k \geq 0} (q-1)^k\binom{x}{k}$ for $x \in \mathbf Z_p$ (as $x \rightarrow 0$).

For $|q-1| < 1$ in $K$ we can think of the series $S(q) := \sum_{n \geq 0} (n)!_q$ as a function of $q$. The OP asked if $S(q)$ is continuous at $q=1$, and we saw it is. It's of course natural, having seen that, to ask if $S(q)$ is continuous in $q$ for all $q$ with $|q-1| < 1$. Let's compare $S(q_1)$ to $S(q_2)$ when $|q_1 - 1| < 1$ and $|q_2-1| < 1$.

For $j \in \mathbf Z^+$, the difference $(j)_{q_1} - (j)_{q_2} = \sum_{k=0}^{j-1} (q_1^k - q_2^k)$ is in $(q_1-q_2)\mathbf Z[q_1,q_2]$, so $|(j)_{q_1} - (j)_{q_2}| \leq |q_1-q_2|$. Therefore in the ring $\mathcal O_K/(q_1-q_2)$ we have $(j)_{q_1} \equiv (j)_{q_2}$ for all $j \in \mathbf Z^+$ (at $j = 0$ too, but that's irrelevant here), so $(n)!_{q_1} \equiv (n)!_{q_2}$ for all $n \in \mathbf Z^+$ and also obviously at $n = 0$. Thus $|(n)!_{q_1} - (n)!_{q_2}| \leq |q_1-q_2|$ for all $n \in \mathbf N$, so
$$
|S(q_1) - S(q_2)| = \left|\sum_{n \geq 0} (n)!_{q_1} - \sum_{n \geq 0} (n)!_{q_2}\right| \leq |q_1-q_2|,
$$
which shows $S(q)$ is uniformly continuous in $q$ for $|q-1| < 1$ in $K$.