Why do we say ‘pairwise disjoint’, rather than ‘disjoint’? I don’t see the ambiguity that ‘pairwise’ resolves. 
Surely if $A$, $B$ and $C$ are disjoint sets then they are pairwise disjoint and vice versa? 
Or am I being dim?
 A: In this context disjoint means $A \cap B \cap C = \emptyset$.
A: Let $A=\{1,2\}, B=\{2,3\},C=\{3,4\}$.  Then the sets are disjoint because $A\cap B\cap C=\emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $A\cap B\not =\emptyset$.
A: Consider the sets $A = \{1,2\}$, $B = \{2,3\}$, $C = \{3, 1\}$.  Then $A\cap B\cap C = \varnothing$, but $A,B,C$ are not pairwise disjoint.
A: $\{1,2\},\{2,3\},\{1,3\}$ are disjoint but not pairwise disjoint. 
A: As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
A: The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
To corroborate my point of view, here is a citation from Halmos:

Pairs of sets with an empty intersection occur frequently enough to justify
  the use of a special word: if $A\cap B=\emptyset$, the sets $A$ and $B$ are
  called disjoint. The same word is sometimes applied to a collection of sets
  to indicate that any two distinct sets of the collection are disjoint;
  alternatively we may speak in such a situation of a pairwise disjoint
  collection.

By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
A: Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
A: This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:

Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]

Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):

"Let $U = U_{\epsilon} = \sum_{j=1}^{\infty} [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."

Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":

if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$

This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
A: If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
