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](https://en.wikipedia.org/wiki/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.