Here is another possible approach, perhaps closer to what the OP had in mind.

Let $S:=[0,1]^2$ be the unit square. "Partition" $S$ naturally into four congruent squares $S_{1,j}$ (with side length $1/2$ each), where $j=1,\dots,4$; the quotation marks are used here because the $S_{1,j}$'s will have some common boundary points. Next, "partition" each $S_{1,j}$ naturally into four congruent squares (with side length $1/2^2$ each), so that we get $4^2$ squares $S_{2,j}$ for $j=1,\dots,4^2$. Continue doing so, so that at the $k$th step we get $4^k$ squares $S_{k,j}$ for $j=1,\dots,4^k$, for each $k=1,2,\dots$.

Take any subset $A$ of $S$. For each $k=1,2,\dots$ and each $j=1,\dots,4^k$, let $$A_{k,j}:=(A\cap S_{k,j})-s_{k,j},$$ where $s_{k,j}$ is the southwest vertex of the square $S_{k,j}$, so that $A_{k,j}\subseteq S_k:=2^{-k}S$.

Suppose that for each $k$ we have a "measure" $D_k$ of dissimilarity for subsets of $S_k$, so that for any two subsets $B$ and $C$ of $S_k$ we have a nonnegative real number $D_k(B,C)$, which is the greater the more "dissimilar" $B$ and $C$ are (and, say, is $0$ if $B=C$); here the term "measure" is used in the general sense, not necessarily in the sense of measure theory. For instance, $D_k(B,C)$ may depend on the Hausdorff distance between $B$ and $C$ or on some "measure" of the symmetric difference of the sets $B$ and $C$ or on some combination thereof.

Then the distance of the set $A$ from uniformity can be defined by the formula $$D(A):=\sum_{k=1}^\infty\frac1{L^k}\sum_{j=1}^{4^k}\sum_{m=1}^{4^k} \frac{D_k(A_{k,j},A_{k,m})}{1+D_k(A_{k,j},A_{k,m})},$$ where $L$ is a real number $>16$ (to ensure the convergence of the series). Then $D(A)$ will be small if, for "most" levels $k$ of "zooming", "most" of the intersections of the set $A$ with all the "$k$-level" small squares $S_{k,j}$ "look similar" to one another. (Of course, $D(A)$ will depend on the choices of $L$ and the dissimilarity "measures" $D_k$.)

Here is another possible approach, perhaps closer to what the OP had in mind.

Let $S:=[0,1]^2$ be the unit square. "Partition" $S$ naturally into four congruent squares $S_{1,j}$ (with side length $1/2$ each), where $j=1,\dots,4$; the quotation marks are used here because the $S_{1,j}$'s will have some common boundary points. Next, "partition" each $S_{1,j}$ naturally into four congruent squares (with side length $1/2^2$ each), so that we get $4^2$ squares $S_{2,j}$ for $j=1,\dots,4^2$. Continue doing so, so that at the $k$th step we get $4^k$ squares $S_{k,j}$ for $j=1,\dots,4^k$, for each $k=1,2,\dots$.

Take any subset $A$ of $S$. For each $k=1,2,\dots$ and each $j=1,\dots,4^k$, let $$A_{k,j}:=(A\cap S_{k,j})-s_{k,j},$$ where $s_{k,j}$ is the southwest vertex of the square $S_{k,j}$, so that $A_{k,j}\subseteq S_k:=2^{-k}S$.

Suppose that for each $k$ we have a "measure" $D_k$ of dissimilarity for subsets of $S_k$, so that for any two subsets $B$ and $C$ of $S_k$ we have a nonnegative real number $D_k(B,C)$, which is the greater the more "dissimilar" $B$ and $C$ are (and is $0$ if $B=C$); here the term "measure" is used in the general sense, not necessarily in the sense of measure theory. For instance, $D_k(B,C)$ may depend on the Hausdorff distance between $B$ and $C$ or on some "measure" of the symmetric difference of the sets $B$ and $C$ or on some combination thereof.

Then the distance of the set $A$ from uniformity can be defined by the formula $$D(A):=\sum_{k=1}^\infty\frac1{L^k}\sum_{j=1}^{4^k}\sum_{m=1}^{4^k} \frac{D_k(A_{k,j},A_{k,m})}{1+D_k(A_{k,j},A_{k,m})},$$ where $L$ is a real number $>16$ (to ensure the convergence of the series). Then $D(A)$ will be small if, for "most" levels $k$ of "zooming", "most" of the intersections of the set $A$ with all the "$k$-level" small squares $S_{k,j}$ "look similar" to one another. (Of course, $D(A)$ will depend on the choices of $L$ and the dissimilarity "measures" $D_k$.)

For instance, for any $L$ and any $D_k$'s we have $D(S)=0$ -- of course, the unit square $S$ is at distance $0$ from uniformity (in itself).

As another example, for the uniform grid $G_n$ (defined in the previous answer) with $n=2^K$ for a natural $K$, any real $L>16$, and any $D_k$'s we have $$D(G_n)\le \sum_{k=K+1}^\infty\frac1{L^k}\,16^k =Cn^{-p}\to0$$ as $n=2^K\to\infty$, where $C:=\dfrac{16}{L-16}$ and $p:=\log_2\dfrac L{16}$. So, we see that $G_n$ is close to uniformity for large $n$, even though the Hausdorff dimension of $G_n$ is $0$ for all $n$ (in big contrast with the Hausdorff dimension of $S$, which is $2$). Thus, again we see that the Hausdorff dimension can hardly have anything to do with the idea of uniformity.