Unless we build a delimiter that is specifically taylored to the input, as in David Carlisle's approach (he uses the token list itself as a delimiter since it is too long to appear within itself), the only safe delimiter is an end-group character token (normally } with TeX's usual catcodes), because unmatched explicit end-group tokens cannot appear in the list of tokens given to \firstofmany. This trailing } can be inserted by expanding \iffalse {\fi #1} from the left, as noted in a couple of answers.
The hard part comes about when trying to insert the corresponding left brace to remove trailing tokens. Several answers simply insert \expandafter\@gobble{\iffalse }\fi after the first item. This has the large drawback that the whole result is not obtained when hitting the \firstofmany function with some \expandafter chains on the left, which means in particular that its result can only be used directly within an \edef or similar expansion, and that it cannot be expanded to be given to another expandable function directly.
My approach is to remove all tokens (except the first item) until a given (mostly arbitrary) marker, then test if the full token list with that part removed only consists in the first item. If it does, then we are done, the token list has only one item, and we leave that as a result. Otherwise, we repeat: keep the first item, remove everything else until the marker, check if what remains is a single item.
It turns out I want blank token lists to give an empty result, which is equivalent to asking for the first item in #1{}. If the argument of \fom@test consists in (optional) spaces, followed by an item, then that is left in the input stream, to be taken as the argument to \unexpanded (it turns out that the item always ends up braced). Otherwise, we call \fom@grab to remove until abc (we know that this marker is present because the initial argument of \fom@test ends with abc), which then calls \fom@test again. The key thing to note is that the leading item is always kept after our functions in the input stream, which means that expanding from the left ("f-expanding") works correctly.
(EDIT: the code was wrong.)
\begingroup
\catcode`@=11
\long\gdef\firstofmany#1{\unexpanded
\iffalse{\fi \fom@grab #1{}abc}}
\long\gdef\fom@test#1{%
\ifcat$\detokenize\expandafter{\fom@gobble#1}$%
\expandafter\fom@i
\else
\expandafter\fom@ii
\fi
{#1}% #1 is {first item}
{\iffalse{\fi \fom@grab #1}}%
}
\long\gdef\fom@grab#1#2abc%
{\expandafter\fom@test\expandafter{\iffalse}\fi{#1}}
\long\gdef\fom@gobble#1{}
\long\gdef\fom@i#1#2{#1}
\long\gdef\fom@ii#1#2{#2}
\endgroup
\long\def\test#1%
{\message{|\unexpanded\expandafter{\romannumeral-`q#1}|}}
\test{\firstofmany{ a bc}}
\test{\firstofmany{ {a\a} bc}}
\test{\firstofmany{ {a\a} abc abc abc}}
\test{\firstofmany{ }}
\csname stop\endcsname
\csname bye\endcsname
\endinput
As an added bonus, this function only requires two steps of expansion to yield its result (similar to many of Heiko Oberdiek's beautiful macros). It is a little bit slower than the naive approach that forbids a specific marker from appearing in the token list.
EDIT2: As Sašo Živanović made me realize, this solution also avoids a common issue: many definitions of \firstofmany fail in the case \halign{#\cr a\firstofmany{b&}\cr}. I got lucky: since the user's token list only appears within braces (sometimes \iffalse{\fi), TeX's alignment mechanism doesn't see it.
