For the expected norm see equation (19) in this paper (or equation (12) in the published version of that paper), which, after some renormalizing, states that $$ \mathbb{E} x_1^{r_1} \cdots x_n^{r_n} = \frac{(n-1)! r_1! \cdots r_n!}{(r+n-1)!}, $$ where $r_i \ge 0$ and $r = r_1 + \cdots r_n$. So in particular (if you can trust my arithmetic) $$ \mathbb{E} ||x||^2 = \frac{2}{n+1}, \quad \mathbb{E} ||x||^4 = \frac{4(n+5)}{(n+1)(n+2)(n+3)}. $$ From these standard Hölder's inequality estimates give that $\mathbb{E} ||x||$ is of order $n^{-1/2}$.

For concentration of the norm, there are general concentration results for convex bodies that apply, in particular "Borell's lemma" which gives much sharper concentration than you asked for — see this answer to another question.

Other relevant results are in this famous paper of Diaconis and Freedman.

For the expected norm see equation (19) in this paper (or equation (12) in the published version of that paper), which, after some renormalizing, states that $$ \mathbb{E} x_1^{r_1} \cdots x_n^{r_n} = \frac{(n-1)! r_1! \cdots r_n!}{(r+n-1)!}, $$ where $r_i \ge 0$ and $r = r_1 + \cdots r_n$. So in particular (if you can trust my arithmetic) $$ \mathbb{E} ||x||^2 = \frac{2}{n+1}, \quad \mathbb{E} ||x||^4 = \frac{4(n+5)}{(n+1)(n+2)(n+3)}. $$ From these standard Hölder's inequality estimates give that $\mathbb{E} ||x||$ is of order $n^{-1/2}$.

For concentration of the norm, there are general concentration results for convex bodies that apply, in particular "Borell's lemma" which gives much sharper concentration than you asked for — see this answer to another question.

Other relevant results are in this famous paper of Diaconis and Freedman.

There are general concentration results for convex bodies that apply (see this answer to another question), but for the simplex you can get away with calculating higher moments directly and using Markov's inequality. For example see equation (19) in this paper (or equation (12) in the published version of that paper), which, after some renormalizing, states that $$ \mathbb{E} x_1^{r_1} \cdots x_n^{r_n} = \frac{(n-1)! r_1! \cdots r_n!}{(r+n-1)!}, $$ where $r_i \ge 0$ and $r = r_1 + \cdots r_n$. So in particular $$ \mathbb{E} ||x||^2 = $$

Other relevant results are in this famous paper of Diaconis and Freedman.

For the expected norm see equation (19) in this paper (or equation (12) in the published version of that paper), which, after some renormalizing, states that $$ \mathbb{E} x_1^{r_1} \cdots x_n^{r_n} = \frac{(n-1)! r_1! \cdots r_n!}{(r+n-1)!}, $$ where $r_i \ge 0$ and $r = r_1 + \cdots r_n$. So in particular (if you can trust my arithmetic) $$ \mathbb{E} ||x||^2 = \frac{2}{n+1}, \quad \mathbb{E} ||x||^4 = \frac{4(n+5)}{(n+1)(n+2)(n+3)}. $$ From these standard Hölder's inequality estimates give that $\mathbb{E} ||x||$ is of order $n^{-1/2}$.

For concentration of the norm, there are general concentration results for convex bodies that apply, in particular "Borell's lemma" which gives much sharper concentration than you asked for — see this answer to another question.

Other relevant results are in this famous paper of Diaconis and Freedman.