Note that the identity matrix is a diagonal matrix with non-negative diagonal elements, hence its singular value decomposition is simply
$$I_n = I_n \cdot I_n \cdot I_n^\ast,$$
all its singular values are $1$, whence $\lVert I_n\rVert_2 = 1$.
am I supposed to use SDV decomposition each time I want to calculate a two-norm?
Not necessarily. Note that an alternative way to define the two-norm is
$$\lVert A\rVert_2 = \sup_{\lVert x\rVert_2 = 1} \lVert Ax\rVert_2.$$
With the singular value decomposition
$$A = U\Sigma V^\ast,$$
we obtain
$$\begin{align}
\sup_{\lVert x\rVert_2 = 1} \lVert Ax \rVert_2 &= \sup_{\lVert x\rVert_2 = 1} \lVert U\Sigma V^\ast x\rVert_2\\
&= \sup_{\lVert x\rVert_2 = 1} \lVert \Sigma V^\ast x\rVert_2\\
&= \sup_{\lVert y\rVert_2 = 1} \lVert \Sigma y\rVert_2,
\end{align}$$
since $U$ and $V^\ast$ are norm-preserving, and it is not hard to see that $\sup_{\lvert y\rVert_2 = 1} \lVert \Sigma y\rVert_2$ is the largest singular value.
This supremum is in general not easy to find, but with that, we can see
$$\lVert A\rVert_2^2 = \sup_{\lVert x\rVert_2 = 1} \lVert Ax\rVert_2^2 = \sup_{\lVert x\rVert_2 = 1} \langle Ax, Ax\rangle = \sup_{\lVert x\rVert_2 = 1} \langle x, A^\ast Ax\rangle,$$
and $A^\ast A$ is a positive semidefinite hermitian matrix, thus $\lVert A\rVert_2^2$ is the largest eigenvalue of $A^\ast A$, which sometimes is easier to compute.
However, computing operator norms like $\lVert A\rVert_2$ is in general not a trivial task.
Regarding the Frobenius norm,
$$\lVert I_n \rVert_F = \sqrt{\sum_{i=1}^n \sum_{j=1}^n \delta_{ij}^2} = \sqrt{n},$$
where $\delta_{ij}$ is the Kronecker symbol, $\delta_{ij} = 1$ if $i = j$, and $\delta_{ij} = 0$ if $i\neq j$.
Best Answer
I assume that the author tries to derive the matrix norms $\|A\|_1$ and $\|A\|_\infty$ induced by vector norms $\|x\|_1$ and $\|x\|_\infty$.
Let's take, for example, $\|\cdot\|_\infty$. We write
$$\|Ax\|_\infty=\sup_i \left|\sum_j A_{ij}x_j\right|\le \sup_i \sum_j |A_{ij}||x_j|\le \sup_j |x_j| \sup_i \sum_j |A_{ij}|.$$ Hence, a good candidate for $\|A\|_\infty$ is $\sup_i \sum_j |A_{ij}|$. We need to prove that this boundary is indeed achieved; it is true, since we can take $i$ where that supremum is achieved and impose $x_j=\mathrm{sign}(A_{ij})$. With such $x$ all our inequalities degenerate to equalities and we conclude that $\|A\|_\infty = \sup_i \sum_j |A_{ij}|$ is a matrix norm induced by $\|x\|_\infty = \sup_j |x_j|$.
The case $\|\cdot\|_1$ is done likewise.