If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expressible in the complete theory of $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$, which is decidable by Tarski's theorem on real-closed fields. For example, quantifying over $n\times 1$ vectors is just $n$ quantifiers over reals (or $2n$ if you want complex numbers).
You mentioned that for each particular $n$, it is decidable by interpreting in the real-closed field, but my point is that this algorithm is uniform in $n$, and so you get a full decision procedure for the multi-sorted logic. That is, given a sentence in the multi-sorted language, we can tell which sorts are quantified over, and so we know how to translate it into a question about real-closed fields, which we can then answer. (I assume that you use a set-up as usual in the multi-sorted logic where each sort gets its own variables and quantifiers.)
If you intend to interpret it over the rationals, then No, since even the $1$-dimensional ring theory of $\langle\mathbb{Q},+,\cdot,0,1,\lt\rangle$ is not decidable, as the integers are definable there, and so you can express the halting problem.
If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expressible in the complete theory of $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$, which is decidable by Tarski's theorem on real-closed fields. For example, quantifying over $n\times 1$ vectors is just $n$ quantifiers over reals (or $2n$ if you want complex numbers).
If you intend to interpret it over the rationals, then No, since even the $1$-dimensional ring theory of $\langle\mathbb{Q},+,\cdot,0,1,\lt\rangle$ is not decidable, as the integers are definable there, and so you can express the halting problem.