You are asking for "a confidence interval the number of people who responded to the survey that will attend". Confidence intervals can be computed for parameters of a model but most people would consider the actual number of people attending an event a random variable itself ( whose distribution might depend on the same parameters as random variables for the survey responses).
Besides the assumption of independence that you already mentioned, you would probably also need to make the assumption that the behavior of different people is identical ( the other i in iid) conditioned on their "attendance".
In this case, one way to model the survey responses for both of the groups separately is with a multinomial model, the natural extension of the binomial model that you mentioned. (There might be other approaches that make use of the ordinal nature of the 5 boxes, see e.g. ordinal regression). In the multinomial model the probability mass function for the numbers, $x$ ($x \in \mathbb{N}^5 \land \sum_i^5 x_i = n$), of $n$ people checking the corresponding of the 5 boxes is given by $p(x | \theta) = \frac{\Gamma(\sum_i x_i + 1)}{\prod_i \Gamma(x_i+1)} \prod_{i=1}^5 \theta_i^{x_i} $ where $\theta\in (0,1)^5 \land \sum_{i=1}^5\theta_i = 1$, i.e. $\theta_i$ is the probability that a person from this group ticks box $i$.
As @whuber pointed out in a comment, if you do not know this probability and don't have data (e.g. form previous years) that link the survey responses to actual attendance there is no point in calculating anything and you start guessing right away.
However if you do know these probabilities for both attenders (A) and non-attenders (Q), since you don't know how many attenders you have nor who the attenders are, you would still need to model them at the same time: $p(x|\theta^A, n_A, \theta^Q, n_Q) = \sum_{\{(x^A, x^Q)| x^A+x^Q=x\land \sum_i^5x^A_i=n^A\land \sum_i^5x^Q_i=n^Q\}} p(x^A|\theta^A)p(x^Q|\theta^Q)$
I don't know if this 10 dimensional summation can be simplified but even not, depending on your data size, it might still be numerically evaluateable (although to costly to compute compute exact confidence intervals).
