Logarithmic Weil height

It depends on what you mean by $|\cdot|$, but probably no.

If by $|\cdot|$ you mean the absolute value on $\mathbb C$, and your algebraic integers are elements of $\mathbb C$, then the answer is no. The logarithmic Weil height of $1-\sqrt3\in\mathbb A^1(\overline{\mathbb Q})$ is $\frac12\log(2)$, which is strictly bigger than $\log(\max(1,|1-\sqrt3|))=0$. (To show that the logarithmic height is $\frac12\log(2)$, use the formula for the logarithmic height on $\mathbb A^1(\mathbb Z[\sqrt3])$ below.)

However, a similar-looking inequality is true if we take account of all possible archimedean norms on the algebraic integers. Here is a precise statement. Let $a_1,\dots,a_n$ be algebraic integers in $\mathbb C$. Then $$ h(a_1,\dots,a_n)\leq\max_{1\leq i\leq n}\left(\max_{\sigma\in G_{\mathbb Q}}\log(|\sigma(a_i)|)\right)\,, $$ where $h$ denotes the logarithmic Weil height on $\mathbb A^n(K)$ and the second $\max$ is taken over all field automorphisms of $\overline{\mathbb Q}$.

This follows relatively straightforwardly from the definition of the Weil height. To cut a long story short, if $K$ is a number field of degree $d$ over $\mathbb Q$, then the Weil height on $\mathbb A^n(\mathcal O_K)$ is given by $$ h(a_1,\ldots,a_n)=\frac1d\sum_{\sigma\colon K\hookrightarrow\mathbb C}\log(\max_{1\leq i\leq n}(|\sigma(a_i)|)) \,, $$ where the sum is taken over all complex embeddings $\sigma\colon K\hookrightarrow\mathbb C$. This is just what you get by specialising the usual formula for the Weil height on $\mathbb P^n(K)$. It follows from this that $$ h(a_1,\dots,a_n) \leq \max_{\sigma\colon K\hookrightarrow\mathbb C}\max_{1\leq i\leq n}\log|\sigma(a_i)| \,, $$ which implies the claimed inequality.

Remark: This all depends, of course, on the chosen normalisation of the Weil height. Since the OP was talking about "algebraic integers", I presumed they were considering Weil heights with the normalisation which leads to a Weil height on $\mathbb P^n(\overline{\mathbb Q})$.