A short answer is "fairly fast" and "not especially big, but more than a fizzle" for the first explosion, followed by a chaotic, and dramatically larger secondary explosion with multiple "centers" that all go off at slightly different times and energies.
The long answer:
Something to remember is the explosion is caused by two reactants consuming one another - matter and antimatter - so basically anything that applies to a similar situation like combustion can also be applied here. The rate of energy release of combustion is determined by the surface area the reactants have in contact - think slowly burning a shot glass of alcohol in air vs. lighting the alcohol-air mist a fire-breather exhales (or sawdust/grain explosions). Google says the surface area of a bowling ball is roughly 232 $in^2$, or (rounding up) 0.15 $m^2$. Air also has a density roughly 1.225 $\frac{kg}{m^3}$, so if we assume all the air within one micrometer of the surface comes into contact with the ball and annihilates with an equal mass of antimatter all at once that will yield approximately 3.3*$10^{10}$ J of energy. That may sound like a lot, and it is, but it's only on the order of $\frac{1}{10,000,000}^{th}$ the total possible energy release, and is roughly $\frac{1}{100}^{th}$ kilotons, i.e. 10 tons of TNT equivalent. That ain't nothin', but it's hardly a city-incinerating nuclear fireball (that also means the total energy is equivalent to about 1,000 megatons, or roughly 10 times bigger than the largest nuclear bomb ever designed).
Further, an explosion that "large" would be enough to blast away any other nearby matter thereby creating a "bubble" of empty space around what's left of the bowling ball. Any further explosive release of energy would have to wait until the air can rush back in and react (this is the principle they use to "blow out" burning oil wells using explosives). On the plus side the eexplosion will also destroy your antimatter bowling ball, breaking it into much smaller chunks and flinging them outward, so when air does get back to them your surface area, and therefore the rate of reaction, will be much higher and you'll get a much larger second BOOM. Also the bubble won't be nearly as big as from a conventional explosive since most of the energy you're releasing will be gamma rays and high energy subatomic particles that won't exert a lot of outward pressure compared to the superheated gasses a traditional explosive yields (less air will be displaced, nor will it be displaced as far).
A final note about the reaction rate; I would not expect matter more than about a micrometer away to be attracted to the bowling ball, even though antimatter and regular matter have opposite electrical charges. Because the anti- atoms and molecules in the bowling ball would be neutral, just like the regular ones in the air there would be no net electrostatic forces over long distances, just short-range IMF's and Van der Waals forces which typically act over a micrometer or less.
As to the time-scale I can only hazard an extremely rough guess given that I don't know much about the spectrum of annihilation products of non-electrons (some will be particles, some will be gamma rays, the particles will be of different types and interact with air in different ways...), but normal air molecules travel at around 500 $\frac{m}{s}$, so the micrometer shell of air would take at minimum (if every molecule in the shell were moving directly towards the ball's center instead of in random directions) about 2 nanoseconds to occur, so the initial "10 ton" detonation could take as little as just under 2 ns (since attraction will speed the molecules up a little). The second, larger explosion would take longer to start, but depends strongly on how large a "bubble" the initial blast creates and the properties of the resulting bowling ball shrapnel (speed, size, number-density, etc).
