I was a bit slow in answering this, but lets take it that the wind mage has some sort of "passive and automatic wind barrier" that isn't effected by the reaction speed of the caster. This seems to have two similar yet different problems in it. Can you slow/stop a bullet, and can you deflect a bullet? Lets look at these separately.
Slowing / Stopping
This problem is all in one straight line, so if we can keep the velocities and accelerations straight, the numbers shouldn't need any calculus. I'll be assuming we're working with a 9mm handgun.
If a wind mage casts a spell at 100 meters directly at the gunman, who has perfect aim, then what kind of wind will be needed to stop the bullet in time?
$force_{wind} = \frac{1}{2} density_{air}*(v_{muzzle}+v_{wind})^2*coef_{drag}*crosssection$
$force_{wind} = m_{bullet}*a_{wind}$
$v_f^2 = v_i^2 + 2a_ {wind}d $
We know roughly that $density_{air} = 1.20458\frac{kg}{m^3}$ , the muzzle velocity of a 9mm is $1200\frac{f}{s} \approx 366\frac{m}{s}$ , the distance between the mage and gun is $100$ meters, the 9mm has a diameter of $9mm$ , which is a cross section of $\frac{9^2\pi} {4}mm ^2$ and that a head on bullet has a drag coefficient of $0.38$ .
Plugging that all in and chugging through, we get $219\frac{m}{s}$ , or $491mph$ . Going by this scale by the National Hurricane Center , 500 mph would make a Category 5 Hurricane look cute, destroying anything in it's path.
Deflecting
This is a different problem, all we need to do is nudge the bullet over by half the width of a human before it gets there.
I'm going to simplify this problem like this: The bullet will travel without air resistance in a perfectly straight line until all the wind acts on it at the same time. This lets us work with triangles instead of curved paths.
The angle of deflection from the bullet to the person to miss is $arctan(\text{The width of the person} / \text{The distance to the person})$
We also know it in terms of velocity $arctan(\text{The speed tangentially away from the person} / \text{The speed towards the person})$
The speed tangentially away from the person, like before is
$force_{wind} = \frac{1}{2} density_{air}*v_{wind}^2*coef_{drag}*crosssection$
$force_{wind} = m_{bullet}*a_{wind}$
But in this case, the wind is acting on the side of the bullet, so the drag is $0.82$ and the cross section is $9mm * 19mm = 171mm^2$ .
$force_{wind} = v_{wind}^2 * 84\mu N$
$a_{wind} = v_{wind}^2 * 0.0113\frac{m}{s^2}$
And our final equation
$\text{human width}/ (\text{total distance} - t*muzzlevelocity) = t *a_{wind} / muzzlevelocity$
$465mm / (100m - t*muzzlevelocity) = t*a_{wind} / muzzlevelocity$
This leads to a quadratic in terms of a_{wind}. We only really need the discriminant to find what the minumum speed of the wind would deflect the bullet, and it's $47\frac{m}{s}$ , or about $104mph$ .
In conclusion, the best defense is a good offense, and if you can already apply hurricane force winds in the direction of a target, as g s pointed out, you probably have better options at that point than trying to blow the bullet out of the air.