The conditional probabilities completely determine the joint probability distribution, at least insofar as it can be revealed by these events, and (obviously) are determined by it. The crux of the matter is to find the set algebra determined by these events $\mathcal X = \{X_1, X_2, \ldots, X_n\}.$
If the $X_i$ aren't already subsets of a common set $\Omega,$ define $$\Omega = \bigcup_{X\in\mathcal X} X,$$ the union of all the events, which is allowed by the axioms of set theory. As a matter of notation, for any $\mathcal{A}\subset \Omega,$ write $\mathcal{A}^0 = \Omega\setminus \mathcal{A}$ for its complement and $\mathcal{A}^1=\mathcal{A}$ for the set itself.
Probability calculations deal with complements of events and unions of disjoint events. We therefore need to construct a collection of sets that is closed under such operations and includes the $X_i.$ To this end construct all $2^n$ possible intersections of the $X_i$ and their complements. That is, for each sequence $\mathbf{e}=(e_1,e_2, \ldots, e_n)$ of zeros and ones, construct
$$\mathcal{X}(\mathbf{e}) = X_1^{e_1} \cap X_2^{e_2} \cap \cdots \cap X_n^{e_n}.$$
If you wish to imagine a Venn diagram of the $X_i$ (in two or more dimensions) the various $\mathcal{X}(\mathbf{e})$ are the connected components of the diagram after the boundaries of the circles (or hyperspheres) have been removed: $e_i=1$ means you're inside $X_i$ and $e_i=0$ means you are outside it.
There might be fewer then $2^n$ distinct nonempty components. (A worked example appears at the end of this post.) Let their number be $N,$ which therefore lies between $0$ and $2^n$ inclusive. These are the atomic sets determined by $\mathcal X.$ Clearly, distinct atomic sets are disjoint.
From now on, assume $\Omega$ is not empty, so that $N \ge 1.$
A probability measure $\mathbb{P}$ assigns non-negative numbers to the $X_i$ (and all sets that can be formed from them and $\Omega$ by means of countable numbers of complements, intersections, and unions). Let this collection of sets generated by $\Omega$ and $\mathcal X$ be called $\mathfrak{S}.$ It is the set of events determined by the $X_i.$ Only events need to be given probabilities.
The values of $\mathbb{P}$ are limited only by two axioms:
$\mathbb{P}(\Omega) = 1.$
When $\mathcal{E}_i \in \mathfrak{S}$ are disjoint, $$\mathbb{P}\left(\bigcup_i \mathcal{E}_i\right) = \sum_i \mathbb{P}(\mathcal {E}_i).$$
Only axiom $(1)$ provides any numerical constraint. Here's the proof. First, note that the intersection of any $\mathcal{E}_i$ in axiom $(2)$ with any of the atomic sets $\mathcal{X}(\mathbf{e})$ is either empty or $\mathcal{X}(\mathbf{e})$ itself. From this it follows that each $\mathcal{E}_i$ is the union of the atomic sets it contains--and those atomic sets, as previously noted, are disjoint. Therefore, the probabilities of the atomic sets determine all the probabilities that can appear in axiom $(2)$.
Because $\Omega$ is the union of all the atomic sets, axioms $(1)$ and $(2)$ together imply the sum of the atomic probabilities is unity. Moreover, the demonstration shows more:
Any event is a union of atomic events. The probability of an event is the sum of the probabilities of the atomic sets within it.
Consequently, if we enumerate the atomic sets once and for all as $\mathcal{A}_1, \ldots, \mathcal{A}_N,$ and write
$$p_i = \mathbb{P}(\mathcal{A}_i)$$
for their probabilities, then
All probability measures on $\mathfrak S$ are determined by vectors $(p_1,p_2\ldots, p_N)$ of non-negative numbers that sum to unity, where $N$ is the number of atomic events determined by the $X_i.$
That imposes one linear equality on $N \ge 1$ independent numbers. Linear algebra tells us the solutions form a manifold of $N-1$ dimensions. Thus,
The number of probabilities that can be independently specified is one less than the number of atomic events determined by $\mathcal X.$
Let's do an example. Within the "universal set" $\Omega=\{0,1,2,3\},$ suppose $X_1 = \{0\},$ $X_2 = \{1,2\},$ and $X_3 = \{0,1,2\}.$ The atomic sets are (in arbitrary order)
$$\mathcal{A}_1 = \{0\},\quad \mathcal{A}_2 = \{1,2\},\quad \mathcal{A}_3 = \{3\}.$$
Consequently there are $2^N=8$ events. They are
$$\eqalign{
\mathfrak{S} &= \{\emptyset,&\{0\}=X_1,&\{1,2\}=X_2,&\{0,1,2\}=X_3,&\{3\},&\{0,3\},&\{1,2,3\},&\{0,1,2,3\}=\Omega\}\\
&= \{\emptyset,&\mathcal{A}_1,&\mathcal{A}_2,&\mathcal{A}_1\cup\mathcal{A}_2,&\mathcal{A}_3,&\mathcal{A}_1\cup\mathcal{A}_3,&\mathcal{A}_2\cup\mathcal{A}_3,&\mathcal{A}_1\cup\mathcal{A}_2\cup\mathcal{A}_3\}.
}$$
Thus $N=3$ and it takes $N-1=2$ numbers to determine a probability measure. You could take them to be, say, $p_1 =\mathbb{P}\left(\{0\}\right)$ and $p_2 = \mathbb{P}(\left(\{1,2\}\right)$ and compute $p_3 = 1 - (p_1+p_2)$ from axiom $(2).$ The axioms force you to choose both $p_i$ to be nonnegative and they cannot sum to any more than $1.$
Finally, note that $\mathbb{P}(X_1)=p_1,$ $\mathbb{P}(X_2)=p_2,$ and $\mathbb{P}(X_3) = p_1+p_2.$ This shows that you cannot necessarily assign probabilities freely to each of the original sets $X_i:$ the axioms may force constraints on them. That's why considerations of the set algebra are unavoidable, because they reveal what those constraints are.
