If $\phi\colon\mathbb S^2\to\mathbb S^2$ is a smooth harmonic map, and we define \begin{align} E(\phi):=\int_{\mathbb S^2}|\nabla \phi|^2dv \end{align} then do we have \begin{align} E(\phi)=8\pi|\deg\phi| \end{align}
Background: I am reading an article, and when the author(s) try to prove corollary 1 from Theorem A&B, I think the above equation is used. And I only calculated the case when $\varphi=\operatorname{id}$, where $\deg\varphi=1$, and the energy is $8\pi$.
The equation is true, and it follows from the several facts about harmonic maps from a surfaces $S$.
First of all, if $\varphi : S\to (N, h)$ is a harmonic map from a Riemann surface $S$ to $(N, h)$, then the Hopf differential is holomorphic. In particular when $S = \mathbb S^2$ is the two sphere, the Hopf differential is zero and thus $\varphi$ is weakly conformal.
When, moreover, that $\varphi :\mathbb S^2 \to (N, h)$ is nonconstant, then $\varphi$ is a branched conformal (minimal) immersion.
Also, when $S$ is a surface and $\varphi$ is a branched conformal immersion, one has $E(\varphi) = 2\operatorname{Area}(\varphi)$, the area with respect to the immersion. This tells you why you got $8\pi$ for the identity map, since $\mathbb S^2$ has area $4\pi$.
The above are well-known. For example, these are more or less stated as facts in section 1 of Sacks-Uhlenbeck's Annals-paper. They are also discussed in the first chapter in Schoen and Yau's book on harmonic maps
Now when $S_1, S_2$ are both surface and $\varphi :S_1\to S_2$ is a branched conformal immersion, then $\varphi$ is an open map. If $S_1$ is compact then $\varphi$ is onto and forms a covering (away from finitely many branch points). The no.of sheet of the cover is exactly (absolute of) the degree of the map $\varphi$. In particular, we have
$$ E(\varphi) =2\operatorname{Area}(\varphi)= 2|\deg\varphi| \operatorname{Area}(S_2).$$