$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$ and is maximised when the covariances take their maximum possible positive values which happens when all the correlations are $+1$.
So the highest variance case for $\sum X_i$ and thus $\bar X$ will be when there is perfect positive correlation between the $X_i$, in which case $\text{SD}(\sum X_i) = \sum \text{SD}(X_i)$ giving $\text{Var}(\bar X) = \frac1{n^2} \text{Var}\left(\sum X_i\right)=\left(\frac1n \sum \text{SD}(X_i)\right)^2$.
$\left(\frac1n \sum \text{SD}(X_i)\right)^2 \le \frac1n \sum \left(\text{SD}(X_i)^2\right) = \frac{1}{n}\sum \text{Var}(X_i)$, as can be shown using the Cauchy–Schwarz inequality, with equality only when all the $\text{SD}(X_i)$ are equal.
So your $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum \text{Var}(X_i)$ is correct,
with equality only when $X_i-E[X_i]=X_j-E[X_j]$ for all $i,j$ so when you have identical variances and perfect positive correlation (though possibly different expectations).