You should consider modeling the situation using the multinomial distribution. I am going to change variables as I would prefer to reserve $n$ for sample size and denote the number of choices by $K$ (i.e., $K$ represents the number of colors, answers etc).
Let $p_k$ be the true proportion of people in the population who would choose the $k^\text{th}$ choice when presented with $K$ choices. You can re-interpret $p_k$ as the probability that a random person would choose the $k^\text{th}$ choice when presented with $K$ choices. Thus, by definition, we have:
$$\sum_{k=1}^K p_k = 1$$
Let $x_k$ stand for the number of people who choose the $k^\text{th}$ object when we sample the choices of $n$ people. Then the density function of ${x_k}$ is given by the multinomial pdf:
$$f(x_1,...x_K|-) = \begin{cases} \frac{n!}{x_1! ... x_K!} p_1^{x_1} ... p_K^{x_K} \quad \text{if} \quad \sum_kx_k=n \\ 0 \quad \text{otherwise}\end{cases}$$
You can then use maximum likelihood theory to estimate $\{p_1,p_2,...p_K\}$ and compute confidence intervals for these estimates.
Computing the confidence intervals would also enable you to compute margin of errors associated with your estimates for a given sample size. These margin of errors will help you calculate the necessary sample sizes to attain a margin of error of 5% with 95% confidence.
MLE, Margin of error and Sample Size Computations
It is not difficult to show that the MLE estimate for $p_k$ is given by:
$$\hat{p}_k = \frac{x_k}{n}$$
The above estimator is unbiased as:
$$E(\hat{p}_k) = \frac{E(x_k)}{n} = \frac{n p_k}{n}=p_k$$
The variance of the estimator is:
$$V(\hat{p}_k) = \frac{V(x_k)}{n^2} = \frac{n p_k (1-p_k)}{n^2}=\frac{p_k (1-p_k)}{n}$$
Assuming that $n$ is sufficiently high, we can use the central limit theorem to approximate the distribution of $\hat{p}_k$ as a normal with the mean at $p_k$ and with variance $\frac{p_k (1-p_k)}{n}$.
Thus, the margin of error is given for a 95% confidence interval is given by:
$$1.96 \sqrt{\frac{p_k (1-p_k)}{n}}$$
We do not know $p_k$ apriori. However, a conservative estimate for $p_k$ would be that it equals $K^{-1}$ (i.e., we assume that all choices are equally likely). The above argument is a bit ad-hoc but perhaps serves the OP's purpose.
Therefore, we have the requirement that:
$$1.96 \sqrt{\frac{K^{-1} (1-K^{-1})}{n}} = 0.05$$
If we let $K=3$ we get the required sample size as $n=341.475$.
PS: The past question on Asymptotic distribution of multinomial seems relevant in the above context and may suggest ways to lend rigor to the above ideas.
