First of all, "cohomological dimension" means two different things. The cohomological dimension of $X$ could be the largest $n$ such that $H^n(X)$ does not vanish. This definition is appropriate for defining the cohomological dimension of a group $G$ to be the cohomological dimension of its classifying space $BG$. It could also mean the supremum of $n$ for all closed subsets $C \subseteq X$ such that $H^n(X,C) \ne 0$. This second definition is more appropriate for geometry. By this second definition, an $n$-ball is $n$-dimensional, because you can let $C$ be its boundary. In this second, more geometric approach, it's standard and better to use Čech cohomology. So I should really write $\check{H}^n(X,C)$, but I won't bother. Also, you can ask for the cohomological dimension with various coefficients, and you might get different answers.

There is a very interesting review by Dranishnikov of cohomological dimension theory of compact metric spaces. It includes his famous construction of a compact metric space whose covering dimension is $\infty$ and whose cohomological dimension (over $\mathbb{Z}$) is 3.

Theorem: (elementary) If $X$ is any topological space and $A$ is an abelian group, then $\dim_A X \le \dim X$, where the right side is the covering dimension.

As Torsten Ekedahl points out, this inequality follows quickly from the definition of Čech cohomology and the fairly similar definition of covering dimension. In both cases you use the nerve of an open covering. In Čech cohomology, you take the limit of the cohomology of the nerve. In covering dimension, you take the lim inf of the dimension of the nerve, QED.

Theorem: (elementary) (1) $\dim_A X \le \dim_{\mathbb{Z}} X$; (2) For every $A$, $\dim_A X = 0$ iff $\dim X = 0$; (3) $\dim_\mathbb{Z} X = 1$ iff $\dim X = 1$; (4) For every $A$ and every $n$-dimensional compact simplicial complex $K$, $\dim_A K = n$.

I would guess that (4) also applies to compact CW complexes.

Theorem: (Alexandroff) If $X$ is a compact metric space and if its covering dimension $\dim X$ is finite, then $\dim X = \dim_{\mathbb{Z}} X$.

Theorem: (Pontryagin) For each prime $p$, there is a "surface" $X$ whose $\mathbb{Z}/p$-dimension is 2, and whose $\mathbb{Z}/q$-dimension is 1 for any prime $q \ne p$.

In light of Alexandroff's theorem, if you want cohomological dimension and covering dimension to differ for a compact metric space, the covering dimension has to be infinite. Whether this was possible was a long-standing open problem; Dranishnikov found the first example. He also describes an example of Dydak and Walsh of a compact metric space $X$ such that $\dim_{\mathbb{Z}} X = 2$ but $\dim_{\mathbb{Z}} X \times X = 3$.