How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then $$ |\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y| $$ so $$ \rho(\alpha x + \beta y) = \rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le \rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y) $$

So it seems we also need $\rho(|x|) =\rho (x)$.

added But $\rho(|x|) =\rho (x)$ can fail for Nowak's definition, this is not the way to do it?

How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then $$ |\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y| $$ so $$ \rho(\alpha x + \beta y) = \rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le \rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y) $$

Note: $\rho(x) = \rho\big(|x|\big)$ from N2.

How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then $$ |\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y| $$ so $$ \rho(\alpha x + \beta y) = \rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le \rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y) $$

So it seems we also need $\rho(|x|) =\rho (x)$.

How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then $$ |\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y| $$ so $$ \rho(\alpha x + \beta y) = \rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le \rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y) $$

So it seems we also need $\rho(|x|) =\rho (x)$.

added But $\rho(|x|) =\rho (x)$ can fail for Nowak's definition, this is not the way to do it?