Closure under intersection with regular sets is natural in the sense that most machine models for language classes are finite-state based with some additional control mechanism, and by some standard constructions a finite automaton describing some other regular language could be coded into this finite-state mechanism to get closure under intersection. This is how most of the proofs that show this closure property work. I will give a more detailed and formal answer under some very specific framework below. But before some remarks on the boolean operations.
1) Note the curious fact that for a trio, closure under intersection is the same as closure under shuffle. The shuffle of two words is the language of all possible interleavings of two words, i.e., if $u,v \in X^*$ for some alphabet $X$, then
$$
u \diamond v := \{ u_1 v_1 u_2 v_2 \cdots u_k v_k \mid k \ge 1, u_i, v_i \in X \cup \{\varepsilon\}\}.
$$
For languages $U, V \subseteq X^*$ we set $U \diamond V = \bigcup_{u \in U, v \in V} u\diamond v$. We can write the shuffle with intersection and the trio operations. Let $\overline X = \{ \overline x : x \in X \}$ be some disjoint copy of $X$ and $h : (X \cup \overline X)^* \to X^*$ the homomorphism given by $h(x) = h(\overline x) = x$ and $h_1, h_2 : (X \cup \overline X)^* \to X$ two homomorphisms given by $h_1(x) = h_2(\overline x) = \varepsilon$, $h_1(\overline x) = x$ and $h_2(x) = x$. Then
$$
U \diamond V = h(h_1^{-1}(U) \cap h_2^{-1}(V)).
$$
This formula also gives that we always have closure under shuffle with regular languages in any trio. Conversely if a trio is closed under shuffle then
$$
U \cap V = h((U \diamond \overline V) \cap (X\overline X)^*)
$$
where $\overline U = \{ \overline u : u \in U \}$ is the image of U under the homomorphism given by $x \mapsto \overline x$.
2) If we allow complementation as already mentioned many interesting and natural language classes are excluded. But this could be stated more formal as was done in Georg Zetzsche's dissertation Monoids as Storage Mechanisms.
If a trio containing only recursively enumerable languages is closed under complementation, then it is precisely the class of all regular languages.
So, if we also want some meaningful way to talk about computability, there is not trio closed under complementation except the regular languages.
3) Let me make my intuitive remarks why closure under intersection with regular languages is a natural operation more precise, or represents a structural property as noted in @babou answer. As said it has to do with the fact that most machine models are finite state based with some additional control (a stack, an infinite tape, a linear bounded tape etc).
One way to make this more precise and unify these models is the notion of valence automata, which is also a central topic in the above mentioned dissertation. A valence automaton is essentially a finite automaton enriched with some ''counting mechanism/or storage mechanism'' in form of a monoid. Imagine you have some additional register that can save an element from some monoid, and you can alter it by multiplication with other monoid elements. The following monoids correspond to the following language classes.
Monoid | Language class
-------------------------------
B | Context-Free
F_2 | Context-Free
B x B | Recursively Enumerable
1 | Regular
Z^n | Blind counter languages
N^n | Partially blind counter languages
where B denotes the bicyclic monoid, F_2 the free group of rank $2$, 1 the trivial monoid, Z the integers, and N the natural numbers, see the mentioned dissertation for more details.
By a slight modification of the product automaton construction we can show that if $U$ and $V$ are accepted by some valence automata over monoids $M$ and $N$, then $U \cap V$ is accepted by some valence automaton over $M \times N$. But if $V$ is regular, we can choose $N = 1$, the trivial monoid. Hence in this case $M \times N \cong M$, so that we have not left our language class, or said differently every language described by valence automata is closed under intersection with regular languages (and in fact they are semi AFL's).
