Okay, that's important knowing the concentration. For an NaOH solution, Normality is equal to Molarity, so your solution has a concentration of 0.2 Molar. One Liter of solution will have 0.2 moles of NaOH in there. Therefore, 3 mL of solution will have 0.0006 moles of NaOH. 6 gallons is equal to about 22.7 liters, so your OH- concentration would be 2.64x10^-5 molar. (That's very dilute). If there was absolutely no acidity in the wine (I.E. if it were just distilled water), your pH would now be about 9.4 which is fairly basic. However, wine is quite acidic with a typical pH being about 4.0. So if we assume that your wine is a typical wine, then the hydrogen ion concentration would be 10^-4. So what we know is that in the 22.7 liters of wine we have 22.7x(10^-4) or 0.00227 moles of H+. OH- will neutralize that H+, so the 0.0006 moles of NaOH you put in there will neutralize the same amount of H+ ions. Your wine will now have 0.00227-0.0006=0.00167 moles of H+.
So now we can figure out the pH of your wine. 0.00167 moles of H+ in 22.7 liters (We're ignoring the paltry 3mL that was added by the volume of NaOH) is equal to a [H+] of 9.74x10^-4 Molar. This means that your pH is now 4.133. You'll never notice that change. So your wine is perfectly safe to drink.