Can we determine a number of objects in a category? Maybe my question is too idealistic, but I try it:
Is there any way to look at a category and be able to tell whether this category has zero, one, countable, infinite or any particular number of objects? If not generally, is it possible in particular cases of categories? And what about determining if the category has/hasnt initial or terminal object?
The example that motivated me for this question: If we have a topological space and take all its compactifications, which form a category - are we able to determine (using category theory tools or whatever else), how many objects are there in the category?
 A: 
Is there any way to look at a category and be able to tell whether this category has zero, one, countable, infinite or any particular number of objects?

No, you have to analyze each category separately. There is no and can't be a general method (as in algorithm) for solving that.

If we have a topological space and take all its compactifications, which form a category - are we able to determine (using category theory tools or whatever else), how many objects are there in the category?

If $X$ is a topological space and $Y$ is any set with a bijection $f:X\to Y$ then $Y$ can be given a topological space structure via $U\subseteq Y$ is open if and only if $f^{-1}(U)$ is open in $X$. In that way $f$ becomes a homeomorphism. And therefore if $K$ is some compactification of some space $X$ then any set $Y$ (equinumerous to $K$) can be turned into a space homeomorphic to $K$, and thus $Y$ can be turned into compactification of $X$. Homeomorphic, but distinct from $K$. Therefore the category of compactifications is at least as big as the category of all sets of a fixed cardinality, which is a proper class.
Even up to homeomorphism compactifications are at least as big as the category of sets. If $K$ is a compactification of some $X$ and $Y$ is any set, then the disjoint union $K\sqcup Y$ together with $\{U |\ U\text{ open in }K\}\cup\{K\sqcup Y\}$ topology is a compact space such that $X$ embeds into it as a dense subset.
The thing becomes interesting if we consider Tychonoff spaces only (the construction above is never Tychonoff). In that situation it can be shown that if $X$ is a Tychonoff space and $K$ its Tychonoff compactification, then $|K|\leq 2^{2^{|X|}}$. And therefore, unlike other situations, Tychonoff compactifications form a set, but only up to homeomorphism. Still, the precise cardinality heavily depends on $X$ and I don't think such classification exists.
Edit: I will show the $|K|\leq 2^{2^{|X|}}$ inequality when $X$ is Tychonoff. Of course if $X$ is finite then it has to be compact discrete and we cannot embed it into any other compact Tychonoff space (as a dense subset), thus $|K|=|X|$.
Now assume $X$ is infinite and consider $\beta X$, the Stone–Čech compactification of $X$. It is known that every compactification of $X$ is an image (even quotient) of $\beta X$. This follows directly from the universal property of $\beta X$: given an embedding $f:X\to K$ into compact $K$ with dense image, we have the induced map $\beta f:\beta X\to K$ whose image is at least $f(X)$. But since $\overline{f(X)}=K$ then $\beta f(\beta X)=\overline{\beta f(\beta X)}=K$ and thus $\beta f$ is onto. Therefore the cardinality of any compactification is at most $|\beta X|$.
Now how is $\beta X$ constructed? One way is to construct it as a certain subset of $[0,1]^C$ where $C=\{f:X\to[0,1]\ |\ f\text{ is continuous}\}$ with the open-closed topology (see the wiki on Stone–Čech). And therefore any compactification is of cardinality at most $\frak{c}^{\frak{c}^{|X|}}$ where $\frak{c}=2^{\aleph_0}$ denotes the cardinality of reals. Thus the cardinality is at most $2^{\aleph_0\cdot 2^{\aleph_0\cdot |X|}}$. But since $X$ is infinite then the above simplifies to $2^{2^{|X|}}$.
Note that $2^{2^{|X|}}$ boundary cannot be improved. It is well known that the Stone–Čech compactification of naturals is of such cardinality. Which of course doesn't answer the question: how many non-homeomorphic compactifications are there for a given $X$?
