I thought about allowing a negative value of $d$ in the definition given in the article in the OP. My definitions do not extend Hausdorff dimension, but considers an alternate interpretation of dimension itself as nonpositive numbers rather than nonnegative numbers.
It seems to me that the definitions of Hausdorff dimension offer little leeway into generalizing into negative dimensions: you would have to admit values of $d$ which are negative in the definition of $\dim_H(X)$, and I don't see that playing very well with the infimum that's stuck in there.
As it stands in the article you linked, we have this $C_H^d$ function (the "Hausdorff content") that takes input of some set in a metric space $X$ and spits out the smallest among the sums of the form $\displaystyle\sum_{i} r_i^d$ where $\{r_i\}$ is some set of real numbers that define the radii of some collection of open balls that cover $D$.
In the article, the value of $d$ is allowed to grow without bound, but to compensate that we took the infimum (to "filter out the big sums"). So, I think it's reasonable that allowing a negative value of $d$ would mean we probably want to look at a function like this
$C_{\tilde{H}}^d = \displaystyle\sup_{d < 0} \left\{ \displaystyle\sum_i r_i^d \right\},$
where $\{r_i\}$ is defined the same way it is for $C_H^d$. This time we use a $\sup$ to compensate for the unboundedness of $d$ toward $-\infty$ (to "filter out the small sums").
Now in the article, they use the function $C_H^d$ to define this other function $\dim_H$ (the Hausdorff dimension) where $\dim_H(D) = \inf\{d \geq 0 \colon C_H^d(D) = 0\}$. Since the $d$'s in the original grew without bound, the filter out the big ones and end up with the infimum.
What we should do is now clear: define the nonpositive Hausdorff dimension
$\dim_{\tilde{H}}(D) = \sup \{d < 0 \colon C_{\tilde{H}}^d(D) = 0 \}.$
Does this result in anything new? Is the nonpositive Hausdorff dimension of a set simply the negative of the positive Hausdorff dimension? Is the Hausdorff dimension of a line (which normally has Hausdorff dimension $1$) simply $-1$ or is it more interesting? Some conjectures that should be investigated to find out:
Conjecture: $\dim_{\tilde{H}}(\emptyset) = 0$?
Conjecture: Theorem (using modified $C_{\tilde{H}}^d$ below): $\dim_{\tilde{H}}(\{1,2,3\}) = 0$
Conjecture: $\dim_{\tilde{H}}(LINE) = -1$?
Conjecture: $\dim_{\tilde{H}}(PLANE) = -2$?
Conjecture: $\dim_{\tilde{H}}(CANTOR SET) = -\frac{\log(2)}{\log(3)}$?
If these conjectures are true, then I suspect that this alternative formulation of dimension is pretty boring in that it doesn't tell us anything we didn't already know. If not, I'm interested in seeing what happens.
Of course this idea could easily generalize to allowing complex-valued $d$'s. We may have to throw a modulus in there, but it could work similarly. In fact we could somehow pick complex-valued dimensions with complex-valued $d$'s or cast it back down to reals while using complex-valued $d$'s and taking moduli. Lots of possibilities there to find something interesting.
addendum:
I took the time to show that $\dim_{\tilde{H}}(\{1\}) = 0$. Before that, while doing this, I decided that we should be using this definition of $C_{\tilde{H}}^d$ (note: I added a negative sign to the previous definition):
$C_{\tilde{H}} = \displaystyle\sup_{d < 0} \left\{ -\displaystyle\sum_i r_i^d \right\}.$
Here we are taking the negative of the sum (why this was important will be apparent soon). For now, set $d=-1$, as it will be illustrative of arbitrary $d<0$. Then,
$\begin{array}{ll} C_{\tilde{H}}^d(\{1\}) &= \displaystyle\sup_{d < 0} \left\{ -\displaystyle\sum_i r_i^d \right\} \\ &= \displaystyle\sup_{d < 0} \left\{ - \frac{1}{r_1} \right\} \\ \end{array}.$
where $r_1$ is the radius of any open ball that covers $\{1\}$. We are seeking the supremum of a set of negative numbers, so we want to make $\frac{1}{r_1}$ as small as possible (in order that $-\frac{1}{r_1}$ be as large as possible), implying that $C_{\tilde{H}}^{-1}(\{1\})=0$. (NOTE: this is why I amended the definition of $C_{\tilde{H}}^d$ -- if we did not put it there, then we would have to deduce $C_{\tilde{H}}^{-1}(\{1\}) = \infty$ by making $r_1$ arbitrarily small! This is a big problem because we will not have any $d$ such that $C_{\tilde{H}}^d(\{1\})=0$ so the dimension will not be defined!)
Now notice that we will get this same behavior no matter which $d \in (-\infty,0)$ we choose, yielding $C_{\tilde{H}}^d(\{1\})=0$. So by definition,
$\dim_{\tilde{H}}(\{1\}) = \sup \{d < 0 \colon C_{\tilde{H}}^d(D) = 0 \} = 0,$
since $0$ is the supremum of $(-\infty,0)$. $\blacksquare$
The same process can be used on the set $\{1,2,3\}$, so I resolve the above conjecture, under the assumption that we use the modified $C_{\tilde{H}}$ (with the negative sign).
Since I am unfamiliar with the techniques of showing the Hausdorff dimension of a line is $1$, I cannot proceed further very easily.