The answer depends on what you mean by "quantities"! As José demonstrated, there are no polynomial absolute invariants, essentially because of homogeneity, however, there are plenty of *rational* ones, obtained as fractions $F/G,$ where $F$ and $G$ are polynomial *relative* invariants of the same weight $k:$

$$ F(g\cdot f)=(\det g)^k F(f),\ G(g\cdot f)=(\det g)^k G(f), \quad g\in GL(V).$$

Your dimension count is one of the common false believes. In fact, sanity is restored when working with rational functions, for

$$ \operatorname{tr\ deg} K(X)^G=\operatorname{tr\ deg} K(X)-\dim O_x,$$

where $X$ is an irreducible algebraic variety over an alg. closed field $K$ of char 0 with an action of an algebraic group $G$, $O_x=G\cdot x$ is a generic orbit, $G_x$ is the stabilizer of $x$, $\dim O_x=\dim G-\dim G_x.$ In fact, rational invariants always separate generic orbits.^{1} What goes wrong with polynomial invariants is that orbits need not be Zariski closed. In your example, due to the presence of dilations, Zariski closure of *every* orbit contains zero. A $G$-invariant polynomial function $F$ is constant on any orbit $O$ and hence its value at any point $x\in O$ is equal to $F(0),$ so $F$ is constant.

Another way to resolve the issue is to replace the group $G$ with its subgroup $[G,G]$, which is the inter-section of the kernels of all 1-dimensional representations of $G$, thus replace $Gl(V)$ with $SL(V)$ om the example. After the $GL(V)$-equivariant identification $Hom(V,V\otimes V)\simeq V^*\otimes V\otimes V$ and polarization, which replaces a homogeneous degree $d$ polynomial function on $W$ with a multilinear map with $d$ arguments $W^\otimes d\to K$, the question reduces to finding multilinear $SL(V)$-invariants. Classical invariant theory shows that they are all obtained by composing $GL(V)$-invariant contractions $V^* \otimes V\to K,$ $\xi\otimes v\mapsto \xi(v)$ and expansions $K\to V^*\otimes V,$ $1\mapsto \sum e_i^*\otimes e_i$, permutations, symmetrizations, antisymmetrizations, and the $SL(V)$-invariant determinants $V^{\otimes k} \to K,$ $v_1\otimes\ldots\otimes v_k\mapsto \det[v_1|\ldots|v_k],$ $k=\dim V.$ Explicit formulas are a bit messy.

Footnotes

^{1}See Vinberg and Popov's article on invariant theory in Algebraic Geometry 4 volume of the yellow Russian Math Encyclopaedia.