Somehow I missed this question when it was originally asked. I'm not entirely sure what *you* mean by a negative bias; the bits you've highlighted in the graph of $M(x)$ aren't when $M(x)$ is negative, but rather where is is, in some sense, "decreasing on average", and I don't know how to formalise that notion. If you're actually interested in the set of $x$ for which $M(x)$ is negative, then it seems clear from the graph that this happens roughly half the time. Here's how to make that notion formal.

In Nathan Ng's paper linked in Micah Milinovich's answer, Ng (very conditionally!) proves the existence of a limiting logarithmic distribution of $M(x)/\sqrt{x}$, so that there exists a measure $\nu$ satisfying
$$\lim_{X \to \infty} \frac{1}{\log X} \int_{1}^{X}{f\left(\frac{M(x)}{\sqrt{x}}\right) \ \frac{dx}{x}} = \int_{\mathbb{R}}{f(x) \ d\nu(x)}$$
for every continuous bounded $f : \mathbb{R} \to \mathbb{R}$; equivalently, for every Borel $B \subset \mathbb{R}$ whose boundary has $\nu$-measure zero,
$$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x)/\sqrt{x} \in B\}}{ \ \frac{dx}{x}} = \nu(B).$$
Furthermore, he calculates the Fourier transform of $\nu$ explicitly and shows, among other things, that $\widehat{\nu}$ is *even* about the origin. This implies that
$$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) < 0\}}{ \ \frac{dx}{x}} = \nu((-\infty,0)) = \nu((0,\infty)) = \lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) > 0\}}{ \ \frac{dx}{x}} = \frac{1}{2}.$$
To show this, we need to know that the set $\{0\}$ has $\nu$-measure zero, but this follows as Ng's explicit formula for $\widehat{\nu}$ is in $L^1(\mathbb{R})$, and hence that $\nu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$. (I don't think Ng actually includes this argument, but it's in my paper that kolik linked in his answer.)

So this shows (conditionally) that the median on $\nu$ is $0$, and hence that $M(x)$ is unbiased, in the sense that the logarithmic density of the set of points where $M(x)$ is negative is the same as that of the set of points where $M(x)$ is positive.

Interestingly, if you consider instead the weighted sum
$$M_{1/2}(x) = \sum_{n \leq x}{\frac{\mu(n)}{\sqrt{n}}},$$
then there *is* a negative bias; this follows from the same methods in my article that kolik linked to, together with the explicit expression
$$M_{1/2}(x) = \frac{1}{\zeta(1/2)} + \sum_{\rho}{\frac{x^{\rho - 1/2}}{(\rho - 1/2) \zeta'(\rho)}} + R(x),$$
where the sum is over the nontrivial zeroes of the Riemann zeta function, and $R(x)$ is some small error term. Other than this, I don't know much about other biases, though it should be the same in number fields, and I recently wrote a paper about this kind of thing in function fields, based on previous work of Byungchul Cha (in which he does not explicitly state the analogous result of there being no bias, though it is clear from the results).

Edit: I forgot to mention Brent and van de Lune's recent paper, where they look at a form of the Lambert series generated by $\mu(n)$ and show that it is negative as $x$ tends to $1$ from below (as juan mentioned in a comment above). But this really isn't telling you anything other than that the Riemann zeta function is negative at $s = 0$, which is a much weaker statement than, say, a pole whose residue is negative (as is the case with the summatory function of the Liouville function).