If we throw $n$ dice. And $X$ is the total number of eyes. Find $\operatorname{var}(X)$.
My idea was to label $X=X_1+\cdots+X_n$ where $X_1$ is the outcome of die $1$ etc.
And because $X_1,\ldots, X_n$ are independent we can write $$\operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n)=n\operatorname{var}(X_1).$$
Am I right? Or can't I break this random variable down like this?
Your method (decomposing variance of a sum into the sum of the variances) and justification (the components of the sum are independent) are correct. This is commonly used to simplify computation of variance.
An important direct consequence of this is that the standard deviation of the sum would increase proportionally to $\sqrt{n}$ for $n$ components in the sum, which means that the standard deviation of the average would decrease in proportion to $\frac{1}{\sqrt{n}}$.