Geometric measure theory and its many elegant definitions give a way to make sense of these notions. Indeed, many of its tools are well suited for this problem of assigning a pointwise meaning to functions.
Let’s work on an open subset $\Omega$ of $\mathbb R^n$. If the limit
$$L_x := \bigcap_{\delta > 0} \operatorname{ess. im}_{B_\delta (x)} f$$
is unique, then you can indeed assign a value canonically to $f(x).$
This is because $f$ is essentially continuous at $x$, a definition from the theory of geometric measure theory. $f$ is said to be essentially continuous at $x$ if there exists a null set $N \subset \Omega$ such that
$$\lim_{y \to x, \, y \in \Omega \setminus N} f(y)$$
exists. Such a limit is necessarily unique if it exists, and course it equals $L_x$ in our case.
One can also look at this from the Lebesgue point perspective. In the setting that $f \in L^\infty (\Omega)$, it is not too surprising that you can canonically assign a value, since if $L_x$ is unique, $f$ is also in the sense of the Lebesgue differentiation theorem, Lebesgue continuous at $x$, in the sense that
$$\lim_{\delta \to 0} \frac{1}{\mu(B_\delta (x))} \int_{B_\delta (x)} |f(y) - L_x| \, dy = 0.$$
Such a limit $L_x$, which exists in fact a.e. for any (locally) $L^1$ function $f$ is known as the sharp representative of $f$ in geometric measure theory, and conditions for its existence at any particular point $x$ are weaker than that which you gave.
