Unusual Result in Kinetics Lab
Your analysis of the expectation is correct, and we can look at it more quantitatively in this way:
Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:

$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$

For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order. 

The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.

For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.

Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.

Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.

For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long. 

The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.  

So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.

UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected. 

1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1