$\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}$ Here's the new question:
$ \SU_2 $ is a subgroup of $ \GL_2(\mathbb{C}) $. $ \PSU_2 $ is not. However $ \PSU_2 \cong \SO_3(\mathbb{R}) $ is a subgroup of $ \GL_3(\mathbb{C}) $.
$ \SU_3 $ is a subgroup of $ \GL_3(\mathbb{C}) $. The smallest $ k $ such that $ \PSU_3 $ is a subgroup of $ \GL_k(\mathbb{C}) $ is $ k=8 $ see the answer by Ycor
In general, I would like to know what is the smallest dimension of a faithful complex representation of $ \PSU_n $?
For any $ n $, $ \SU_n $ has dimension $ n^2-1 $ so the adjoint representation realizes $ \PSU_n $ as a subgroup of $ \SO_{n^2-1}(\mathbb{R}) $ which in turn is contained inside of $ \GL_{n^2-1}(\mathbb{C}) $.
My intuition is that the adjoint representation $ n^2-1 $ should always be the smallest dimension containing $ \PSU_n $ as a matrix subgroup. However I know this is not true for all simple groups. For example the exceptional group $ G_2 $ has dimension $ 14 $ but already has a faithful representation in dimension $ 7 $.
If $ \SU_n $ is inside of $ \GL_k $ then it must be in the maximal compact subgroup $ \U_k $ and in fact in the maximal semi-simple compact $ \SU_k $. So the question really becomes what is the least $ k $ such that $ \PSU_n $ is a subgroup of $ \SU_k $?