Let $f$ be an analytic isomorphism on the unit disc $D$, find the area of $f(D)$

Applying Parseval's identity (see e.g. Proof of Parseval's identity) to $$ f'(z) = \sum_{n=0}^\infty (n+1)a_{n+1} z^n $$ gives $$ \int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, d\theta = 2 \pi \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n} $$ for $0 \le r < 1$. If $f$ is injective on the unit disk $D$ then $$ \text{area}\, f(D) = \int_{0}^{1} \int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, r d\theta dr \\ = 2 \pi \int_{0}^{1} \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n+1} \, dr \\ = \pi \sum_{n=0}^\infty (n+1) \lvert a_{n+1} \rvert^2 = \pi \sum_{n=1}^\infty n \lvert a_{n} \rvert^2 \, . $$

Exchanging the order of summation and integration can be justified for example by the monotone convergence theorem.

(Note that the factor $\pi$ is missing in the formula in your question.)