A die is rolled 6 times; what is the chance that the first roll is a one or the last roll is a one. This question appears on page 244 of "Statistics, 4th ed" by David Freedman.

The text of the question is:

A die is rolled 6 times. 
The chance that the first roll is an ace or the last roll is an ace equals ______. 
Answer choices are: $(1/6 + 1/6)$, $(1/6 * 1/6)$, and "neither of these".

The correct answer, according to the book is "neither of these".

I tried calculating the probability assuming that any number of aces can appear in the six rolls but that the first and the sixth rolls must result in aces.

I got ~0.278 as the answer. 

$(1/6)^2*(5/6)^4 + (1/6)^3*(5/6)^3*4!/3! + (1/6)^4*(5/6)^2 * 4! / (2!*2!) + (1/6)^5*(5/6)*4!/3! + (1/6)^6$

Is this correct?

My question is: how can one infer from this question that we are to consider all cases of an "ace" appearing on any of the six rolls. Initially, I thought I could consider the six rolls as independent.
You are correct that the six rolls are independent. So the probability that the first roll is an ace is $1/6$, and similarly, the probability that the last roll is an ace is $1/6$. However, the probability of either event happening is not just the sum: you need to subtract the probability of their intersection. The probability of both the first and last roll being aces is $1/6^2$, so the final answer is $1/6 + 1/6 - 1/6^2 = 11/36$.
The probability that the first roll is a 1 is 1/6.  The probability that the last roll is a 1 is 1/6.  Since you're looking at the first OR the last role is a 1, you add the two probabilities.  So the answer is 1/6 + 1/6 = 2/6 = 1/3.