I am aware that for a discrete random variable X with equally likely outcomes, we have:
$$E[X] = \frac{1}{n}\sum_{i=1}^n x_i$$
Can we make the claim that for two not independent discrete random variables X and Y with equally likely outcomes, the following hold true:
$$E[XY] = \frac{1}{n}\sum_{i=1}^n x_iy_i$$ and
$$E[X^2Y^2] = \frac{1}{n}\sum_{i=1}^n x_i^2 y_i^2$$
If not, can you please someone provide some guidelines to compute $E[XY]$ and $E[X^2Y^2]$?
Your question is unclear. What do the $x_i$ and $y_i$ represent?
The only reasonable interpretation I can make is that $X$ is a random variable with discrete support $\mathcal X = \{x_1, x_2, \ldots, x_n\}$ such that for each $i \in \{1, 2, \ldots, n\}$, $$\Pr[X = x_i] = \frac{1}{n};$$ that is to say, $X$ is discrete uniform on the set $\mathcal X$; similarly, $Y$ is discrete uniform on the set $\mathcal Y = \{y_1, y_2, \ldots, y_n\}$. Then $$\operatorname{E}[X] = \sum_{i=1}^n x_i \Pr[X = x_i] = \frac{1}{n} \sum_{i=1}^n x_i,$$ as you claimed; but $$\operatorname{E}[XY] = \sum_{i=1}^n \sum_{j=1}^n x_i y_j \Pr[(X = x_i) \cap (Y = y_j)] \overset{\text{ind}}{=} \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n x_i y_j = \frac{1}{n} \sum_{i=1}^n x_i \sum_{j=1}^n y_j,$$ where $\overset{\text{ind}}{=}$ requires that $X$ and $Y$ be independent. If not, then you would need to know the joint distribution of $(X,Y)$, because uniformity of the marginal probabilities is not sufficient to imply independence (e.g., there could be perfect association).
So for example, suppose $\mathcal X = \{1, 3, 5, 7\}$ and $\mathcal Y = \{2, 4, 6, 8\}$, with $n = 4$, and each of $X$ and $Y$ are uniformly distributed on these sets, respectively. Then in the case of independence, $\Pr[(X = 1) \cap (Y = 2)] \overset{\text{ind}}{=} \Pr[X = 1]\Pr[Y = 2] = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$, but if for example $$Y = X + 1,$$ then $\Pr[(X = 1) \cap (Y = 2)] = \frac{1}{4}$, not $\frac{1}{16}$.