We know that if $A$, $B$ and $C$ are mutually disjoint events i.e. if $A \cap B = B \cap C = C \cap A = \emptyset$, then $$\mathbb{P}(A \cup B \cup C) = P(A) + P(B) + P(C)$$
Now consider two events $E_1$ and $E_2$ that are not mutually disjoint events i.e. if $E_1 \cap E_2 \neq \emptyset$ and we want to evaluate $P(E_1 \cup E_2)$.
From the figure, it is apparent that the intersection $E_1 \cap E_2$ is counted twice: once as part of $E_1$ and once as part of $E_2$. Hence, it needs to be subtracted off. For a more rigorous derivation, see below the figure.
The idea is to split $E_1 \cup E_2$ into three disjoint sets as follows. $$E_1 \cup E_2 = (E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1)$$
Note that $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ and $(E_2 \backslash E_1)$ are mutually disjoint sets. Hence, we have that
$$P(E_1 \cup E_2) = P((E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1))\\ = P(E_1 \backslash E_2) + P(E_1 \cap E_2) + P(E_2 \backslash E_1)$$
Call the above equation $\star$.
Now note that $$(E_1 \backslash E_2) \cup (E_1 \cap E_2) = E_1$$
Since $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ are mutually disjoint sets, we have that $$P(E_1 \backslash E_2) + P(E_1 \cap E_2) = P(E_1)$$
Hence, $$P(E_1 \backslash E_2) = P(E_1) - P(E_1 \cap E_2)$$
Similarly, since $$(E_1 \cap E_2) \cup (E_2 \backslash E_1) = E_2$$ are mutually disjoint sets, we have that $$ P(E_1 \cap E_2) + P(E_2 \backslash E_1) = P(E_2)$$
Hence,
$$ P(E_2 \backslash E_1) = P(E_2) - P(E_1 \cap E_2)$$ Now plug in for $P(E_1 \backslash E_2)$ and $P(E_2 \backslash E_1)$ in $\star$, to get what you want.
One of the axioms of probability is that if $A_1, A_2, \dots$ are disjoint, then
$$\begin{align}
\mathbb{P}\left(\bigcup_{i=1}^{\infty}A_i\right) = \sum\limits_{i=1}^{\infty}\mathbb{P}\left(A_i\right)\text{.}\tag{*}
\end{align}$$
It so happens that this is also true if you have a finite number of disjoint events. If you're interested in more detail, consult a measure-theoretic probability textbook.
Let's motivate the proof for the probability of the union of three events by using this axiom to prove the probability of the union of two events.
Theorem. For two events $A$ and $B$, $\mathbb{P}\left(A \cup B\right) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$.
Proof. Write $$A \cup B = \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right)\text{.}$$ Notice also that $A = \left(A \cap B^{c}\right) \cup\left(A \cap B\right)$ and $B = \left(B \cap A^{c}\right) \cup \left(A \cap B\right)$.
Since we have written $A$ and $B$ as disjoint unions of sets, applying (*) in the finite case, we have that
$$\begin{align}
\mathbb{P}\left(A\right) &= \mathbb{P}\left(A \cap B^{c}\right) + \mathbb{P}\left(A \cap B\right) \\
\mathbb{P}\left(B\right) &= \mathbb{P}\left(B \cap A^{c}\right) + \mathbb{P}\left(A \cap B\right) \\
\end{align}$$
Similarly, since $A \cup B = \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right)$ is a disjoint union of sets,
$$\begin{align}
\mathbb{P}\left(A \cup B\right) &= \mathbb{P}\left[ \left(A \cap B\right) \cup \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right) \right] \\
&= \overbrace{\mathbb{P}\left(A \cap B\right) + \mathbb{P}\left(A \cap B^{c}\right)}^{\mathbb{P}(A)} + \mathbb{P}\left(A^{c} \cap B\right) \\
&= \mathbb{P}\left(A\right) + \overbrace{\mathbb{P}\left(A^{c} \cap B\right)}^{\mathbb{P}(B)-\mathbb{P}(A \cap B)} \\
&= \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right)\text{. } \square
\end{align}$$
Now, armed with the result that we proved in the previous theorem, we can now prove the result for the probability of the union of three events.
Theorem. $\mathbb{P}\left(A \cup B \cup C\right) = \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) + \mathbb{P}\left(C\right) - \mathbb{P}\left(A \cap B\right) - \mathbb{P}\left(A \cap C\right) - \mathbb{P}\left(B \cap C\right) + \mathbb{P}\left(A \cap B \cap C\right)$
Proof. Since $A \cup B \cup C = \left(A \cup B\right) \cup C$, by the previous theorem,
$$\begin{align}
\mathbb{P}\left(A \cup B \cup C\right) &= \mathbb{P}((A \cup B)\cup C) \\
&= \overbrace{\mathbb{P}\left(A \cup B\right) + \mathbb{P}\left(C\right) - \mathbb{P}\left[\left(A \cup B\right) \cap C\right]}^{\text{applying the previous theorem to }\mathbb{P}((A \cup B)\cup C)} \\
&= \overbrace{\mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right)}^{\mathbb{P}\left(A \cup B\right) \text{ from the previous theorem}} + \mathbb{P}\left(C\right) - \mathbb{P}\left[\overbrace{\left(A \cap C\right) \cup \left(B \cap C\right)}^{(A \cup B)\cap C\text{ (distributive property of sets)}}\right] \\
&= \mathbb{P}\left(A\right) + \mathbb{P}\left(B\right) - \mathbb{P}\left(A \cap B\right) + \mathbb{P}\left(C\right) \\
&\qquad- \overbrace{\Big[\mathbb{P}\left(A \cap C\right) + \mathbb{P}\left(B \cap C\right) - \mathbb{P}\left[\left(A \cap C\right) \cap \left(B \cap C\right) \right]\Big]}^{\text{applying the previous theorem to }\mathbb{P}\left(\left(A \cap C\right) \cup \left(B \cap C\right)\right)}\text{,}
\end{align}$$
and since $\left(A \cap C\right) \cap \left(B \cap C\right) = A \cap B \cap C$, the result is proven. $\square$
Best Answer
The minimum would be if they were all the same event, in which case $P(A \cup B \cup C) = \frac 12$. The maximum would be if they covered the sample space, in which case $P(A \cup B \cup C) = 1$. The only bounds you can really get are $$P(A \cup B \cup C) \ge \max(P(A),P(B),P(C))$$ and $$P(A \cup B \cup C) \le P(A) + P(B) + P(C),$$ besides the obvious $P(A \cup B \cup C) \le 1$.