Suppose we have a continuous function $f: U \to \mathbb{R}$ where $U$ is an unbounded open subset of $\mathbb{R}^n$. For every sequence $\{x_n\} \subset U$ converging to the boundary $x \in \partial U$, we have $f(x_j) \to +\infty$. Note here I am not assuming the divergence to infinity to be uniform. Furthermore, if $\{x_j\} \subset U$ and $\|x_j\| \to \infty$, $f(x_j) \to +\infty$.
Now if we define $\hat{f}: \mathbb{R}^n \to \mathbb{R} \cup \{+\infty\}$ by \begin{align*} \hat{f} = \begin{cases} f(x), \qquad \text{ if } x \in U, \\ +\infty, \qquad \text{ if } x \notin U. \end{cases} \end{align*} Is $\hat{f}$ continuous in the usual extended topology? I feel like the answer should be yes since sequentially $\hat{f}$ is continuous at every point. But since I am only assuming the divergence at the boundary to be sequential not uniform, I am not sure whether there are some subtleties I left out.
Yes, $\hat{f}$ is continuous. As you say, it is sequentially continuous, and that is all that is needed for continuity for a function on $\mathbb{R}^n$ (or on any metric space).
It is not even necessary to stipulate that $f(x_j)\to+\infty$ when $\|x_j\|\to\infty$. You only need $\hat{f}$ to respect sequences which actually converge in $\mathbb{R}^n$.