Haskell - 366 351 344 337 333 characters
(One line break in main added for readability, and no line break needed at end of last line.)
import Data.List
import Data.Char
l=length
t=filter
m=map
f c|isAlpha c=toLower c|0<1=' '
h w=(-l w,head w)
x!(q,w)='|':replicate(minimum$m(q?)x)'_'++"| "++w
q?(g,w)=q*(77-l w)`div`g
b x=m(x!)x
a(l:r)=(' ':t(=='_')l):l:r
main=interact$unlines.a.b.take 22.sort.m h.group.sort
.t(`notElem`words"the and of to a i it in or is").words.m f
How it works is best seen by reading the argument to interact backwards:
map f lowercases alphabetics, replaces everything else with spaces.words produces a list of words, dropping the separating whitespace.filter (notElemwords "the and of to a i it in or is") discards all entries with forbidden words.group . sort sorts the words, and groups identical ones into lists.map h maps each list of identical words to a tuple of the form (-frequency, word).take 22 . sort sorts the tuples by descending frequency (the first tuple entry), and keeps only the first 22 tuples.b maps tuples to bars (see below).a prepends the first line of underscores, to complete the topmost bar.unlines joins all these lines together with newlines.
The tricky bit is getting the bar length right. I assumed that only underscores counted towards the length of the bar, so || would be a bar of zero length. The function b maps c x over x, where x is the list of histograms. The entire list is passed to c, so that each invocation of c can compute the scale factor for itself by calling u. In this way, I avoid using floating-point math or rationals, whose conversion functions and imports would eat many characters.
Note the trick of using -frequency. This removes the need to reverse the sort since sorting (ascending) -frequency will places the words with the largest frequency first. Later, in the function u, two -frequency values are multiplied, which will cancel the negation out.
