First, I'd like to revisit Will Sawin's upper bound, but I make no promises about off-by-one errors (which I think I made in comments elsewhere on this page).  The product of any sequence of $x$ consecutive integers has $p$-valuation at least $\lfloor \frac{x}{p-1}\rfloor - \lfloor \log_p(x+1) \rfloor$.  The $p$-valuation of $\binom{n}{a}$ is the number of carries when adding $a$ to $n-a$ in base $p$, so this is bounded above by $\lfloor \log_p n \rfloor$.  In other words, the product of the $v = b-(n-a)$ integers from $n-a+1$ to $b$ must satisfy $\lfloor \frac{v}{p-1}\rfloor - \lfloor \log_p(v+1) \rfloor \leq \lfloor \log_p n \rfloor$.  Pushing some terms around yields the bound $$v \leq (p-1)\lfloor log_p(n) \rfloor +(p-1) \lfloor log_p log_p n \rfloor + 2p - 2.$$

In particular, $v \leq \lfloor log_2(n) \rfloor + \lfloor log_2 log_2 n \rfloor$.  If we only worry about making $\frac{n!}{a!b!}$ into a 2-adic integer instead of an ordinary integer, then this bound is sharp, since the case $n=2^k, a=2^k-1, b = 2^y-1$ saturates it when $k = 2^y-y-1$.

Incidentally, there are infinitely many integers $n$ for which the virtue $v = a+b-n$ is bounded above by 1.  In particular, any Mersenne number $n = 2^k-1$ satisfies $\binom{n}{a} \equiv 1 \pmod 2$, so we get a 2-adic obstruction to larger virtue.

When $n$ is large, $(p-1) \lfloor \log_p(n) \rfloor > \lfloor \log_2(n) \rfloor$ for odd $p$, so this suggests that it is good to check 2-adic valuations when eliminating candidate triples.  Following Noam D. Elkies's suggestion, I rewrote my program in C.  Also, I found an overflow bug in my valuation routine, as a result of using 32 bit integers and multiplying unnecessarily, so some of the previous results for $n>65536$ were false.  You can find the original SAGE code by looking at the edit history of this answer.  Since the C code is somewhat longer than the SAGE version, I put it [here on my web page][1].  

Anyway, starting with k=3 and running up to $2^{25} = 33554432$ took about 48 hours on a 12-core desktop.  Here are the results in the form $a, b, n, v$:

    3, 5, 6, 2
    6, 7, 10, 3
    11, 29, 36, 4
    14, 47, 56, 5
    47, 59, 100, 6
    59, 110, 162, 7
    23, 241, 256, 8
    31, 239, 261, 9
    187, 239, 416, 10
    447, 620, 1056, 11
    239, 1853, 2080, 12
    1439, 1775, 3201, 13
    1663, 2735, 4384, 14
    14335, 18458, 32778, 15
    9209, 23615, 32808, 16
    13106, 52447, 65536, 17
    26207, 39347, 65536, 18
    34719, 227444, 262144, 19
    109237, 152927, 262144, 20
    59039, 465398, 524416, 21
    230111, 294839, 524928, 22
    496123, 3698204, 4194304, 23
    1007871, 3186457, 4194304, 24
    983546, 7405087, 8388608, 25
    1029947, 7358687, 8388608, 26
    527036, 33027423, 33554432, 27
    2479487, 31074973, 33554432, 28

The last few entries are reasonable evidence that I don't lose much by only considering powers of two when computing optimal virtue.  I computed up to $n=2^{39} = 549755813888$, and have $n=2^{40}$ in progress - this last computation will take about 42 hours.  Unlike the previous case, I didn't bother to precompute valuations, due to memory constraints.  [Code is here][2] - it has no input checks, so bad inputs will produce a segfault.

    2, 3, 4, 1
    2, 7, 8, 1
    5, 14, 16, 3
    9, 26, 32, 3
    19, 49, 64, 4
    17, 116, 128, 5
    23, 241, 256, 8
    53, 467, 512, 8
    314, 719, 1024, 9
    482, 1574, 2048, 8
    1943, 2164, 4096, 11
    1979, 6227, 8192, 14
    3317, 13081, 16384, 14
    6548, 26234, 32768, 14
    26207, 39347, 65536, 18
    56375, 74354, 131072, 17
    109237, 152927, 262144, 20
    58559, 465749, 524288, 20
    117350, 931247, 1048576, 21
    511613, 1585561, 2097152, 22
    1007871, 3186457, 4194304, 24
    1029947, 7358687, 8388608, 26
    839773, 15937469, 16777216, 26
    2479487, 31074973, 33554432, 28
    28499711, 38609183, 67108864, 30
    13640318, 120577439, 134217728, 29
    26335607, 242099879, 268435456, 30
    149450463, 387420479, 536870912, 30
    42571871, 1031169982, 1073741824, 30
    210421817, 1937061863, 2147483648, 32
    808182959, 3486784371, 4294967296, 34
    66684221, 8523250405, 8589934592, 34
    57373109, 17122496111, 17179869184, 36
    1071383543, 33288354863, 34359738368, 38
    194490575, 68524986199, 68719476736, 38
    722568911, 136716384599, 137438953472, 38
    2908726651, 27196918033, 274877906944, 40
    5820696123, 543935117807, 549755813888, 42
    275025321599, 824486306219, 1099511627776, 42 (?)

The optimal virtue for $n=2^k$ hews quite close to the 2-adic upper bound, but some kind of strong height result would be necessary to find enough structure in the solutions to get a guaranteed lower bound for large $n$.

  [1]: http://www.math.tsukuba.ac.jp/~carnahan/MO2.c
  [2]: http://www.math.tsukuba.ac.jp/~carnahan/MO2a.c