The conjugate of $f(x)=\|y\|_1$ is, by definition, $$ f^*(z) = \sup_y \{y^Tz - \|y\|_1\} $$
Also we can write $$f(y)=\|y\|_1 = \max_{\|p\|_\infty\leq 1} y^Tp $$
By using this, we can get the conjugate of $f$: $$ f^*(z) = \begin{cases}0 & \text{ if } \|z\|_\infty \leq 1\\ \infty & \text{otherwise} \end{cases} $$
Am I right? Moreover, if $g(x)=\alpha\|y\|_1$, how to get the conjugate of $g$? Should the dual variable change to $z/\alpha$?
Yes, you're doing good. For the convex conjugate (aka Legendre transform) of an arbitrary norm, see this answer.
Regarding the second part of your question,
Claim: Let $X$ be a Hilbert space, $\alpha$ be a nonzero real number, $f : X \rightarrow (-\infty, +\infty]$ be a function, and $g = \alpha f$. Then $g^*(x) \equiv \alpha f^*(x/\alpha)$.
Proof. For any $x \in X$, we have \begin{equation} g^*(x) := \sup_{z \in X}\langle x, z\rangle - g(z) = \sup_{z \in X}\langle x, z\rangle - \alpha f(z) = \alpha \sup_{z \in X}\langle x/\alpha, z\rangle - f(z) =: \alpha f^*(x/\alpha). \end{equation}
Thus in particular, taking $f = \|.\|_1$, we obtain $(\alpha \|.\|_1)^*(x) = \alpha \|x/\alpha\|^*_1 = \begin{cases}0, &\mbox{ if }\|x\|_\infty \le \alpha,\\+\infty, &\mbox{ otherwise.}\end{cases}$.
For more transformation rules concerning convex conjugates, you may checkout this answer. Another useful resource is this table.
