Let $$ X_t = \beta_1 + \beta_2 t + W_t + \theta W_{t-1} $$ where $t \in \mathbb{Z}$ and $W_t \sim N(0, \sigma_W^2)$ i.i.d. I need to show that $\overline{X}_n = n^{-1}\sum_{t=1}^nX_t$ is asymptotically normal.
My attempt: Well this would have been easy if $(X_t)$ is i.i.d. This is not the case I believe because $\mathbb{E}[X_t] = \beta_1 + \beta_2 t$ and $\operatorname{Var}(X_t) = (1 + \theta^2)\sigma_W^2$. And so $$ \overline{X}_n = \frac{1}{n}\sum_{t=1}^nX_t \sim N\left(\beta_1 + \beta_2 n^{-1} \sum_{t=1}^nt, \frac{1}{n}(1 + \theta^2)\sigma_W^2 \right) \sim N\left(\beta_1 + \beta_2\frac{n-1}{2}, \frac{1}{n}(1 + \theta^2)\sigma_W^2 \right). $$ Why would this sample mean converge to a normal distribution?
You are correct that $(X_t)$ is not iid. $$Cov(X_t,X_{t-1}) = Cov(\beta_1 + \beta_2 t + W_t + \theta W_{t-1}, \beta_1 + \beta_2 t + W_t + \theta W_{t-1})$$ $$= Cov(\theta W_{t-1},W_{t-1}) = \theta \sigma_W^2$$
But this covariance also means you can't work out $Var(\bar{X})$ just using the values of $Var(X_t)$. $Var(X+Y) = Var(X) + Var(Y)$ only holds when $Cov(X,Y)=0$. You also made some algebraic slips: $\frac1n \sum_{t=1}^n t = \frac{n+1}2$, not $\frac{n+1}2$, and if the $x_t$ did have $0$ covariance, you would need to divide the sum of their variances by $\frac1{n^2}$ to get $Var(\bar{X})$.
The trick here is that while the $X_t$ aren't iid, the $W_t$ are, so we want to rewrite the problem in terms of the $W_t$. $$\bar{X}=\frac1n \sum_{t=1}^n X_t=\frac1n \sum_{t=1}^n (\beta_1 + \beta_2 t + W_t + \theta W_{t-1})$$ $$= \beta_1 + \frac{n+1}2 \beta_2 + \frac1n \theta W_0 + \frac1n \sum_{t=1}^{n-1}(1+\theta)W_t + \frac1n W_n$$ From here, you can use the independence and normality of the $W_t$ to derive the distribution, like you were trying to do with the $X_t$.
Also, the word 'asymptotically' in the question is a bit odd, as $\bar{X}$ is normal already for $n$, without considering asymptotics, and the mean increases with $n$, so $\bar{X}$ can't converge in distribution.