The following picture easily generalizes to show that $f(n)$ asymptotically approaches $1/n$.

The two triangles of area $1/5$ can clearly each be made larger at the expense of the other triangles, in particular decreasing the two of area $21/100$, until those four triangles have equal area. How this compares to Willie Wong's estimates, I haven't worked out. (In case it's not clear, for $n=2k+1$, the generalization has $k$ equal triangles generalizing the two of size $21/100$ and $k-1$ generalizing the one of size $9/50$.)

**Added 6/12/13**: Aaron Meyerowitz has shown that the construction I gave above is asymptotically kind of crummy compared to the approach described by Tapio Rajala and Willie Wong, and certainly worse than what's implied by the discrepancy results in the paper by Bernd Schulze linked to by Gregor Samsa and Günter M. Ziegler. Its excess over $1/n$ is $O(1/n^2)$, as opposed to $O(1/n^3)$. (Even before the quadratic tweaking to make the largest triangles as small as possible, they have area $(1-4/n^2)/(n-1)$, which is also $O(1/n^2)$ away from $1/n$.)

For $n=5$, however, the construction I gave does give a smaller largest triangle than the alternative, with area

$${\sqrt2-1\over2}\approx0.2071\ldots < {\sqrt{13}-1\over12}\approx0.2171\ldots $$

according to Aaron. (Indeed, even the untweaked area, $21/100$, is smaller.) So I wonder:

Does my (tweaked) construction give the smallest
largest triangle for $n=5$?

Note, however, that my (tweaked) construction for $n=5$ does *not* minimize the *discrepancy*. That is, while it may minimize the size of the largest triangle for $n=5$, it does not minimize the difference between the largest *and smallest* triangle. When the four large triangles have area $(\sqrt2-1)/2$, the small triangle has area $3-2\sqrt2$, so the discrepancy is

$${\sqrt2-1\over2}-(3-2\sqrt2) = {5\sqrt2-7\over2}\approx0.0355\ldots \gt 0.03 = {21\over100}-{9\over50}. $$

(I.e., the tweaking made the largest triangle a little bit smaller, but it made the smallest triangle a *lot* smaller.) My reason for mentioning this is to illustrate the following:

While minimizing discrepancy has
obvious implications for minimizing the size of
the largest triangle, the two problems
are not the same.

To be sure, it's possible this point has already been made. In particular, I haven't looked through the Schulze paper closely. But I thought it might be worth making explicit.