This is not true for $\text{GL}(V)$, but there is a salvage. As a $\text{GL}(V)$-representation, the decomposition of $V^{\otimes n}$ is described by Schur-Weyl duality, and in particular it is a direct sum of Schur functors $S^{\lambda}(V)$. The Schur functors applied to $V$ describe all polynomial $\text{GL}(V)$-representations (those on which $M \in \text{GL}(V)$ acts by matrices with entries polynomial in the entries of $M$). (I need some assumptions on the underlying field; I think characteristic zero suffices?)
Non-polynomial representations include, for example, $M \mapsto |\det(M)|^r$ for real $r$ if the base field is $\mathbb{R}$ or $\mathbb{C}$; these can never occur as a subrepresentation of $V^{\otimes n}$. In other words, we really want to think about $\text{GL}(V)$ as an algebraic group here.
For finite groups we have the following classical exercise.
Theorem: Let $G$ be a finite group and $V$ a faithful complex representation of $G$. Then every irreducible representation of $G$ occurs as a subrepresentation of $V^{\otimes n}$ for some $n$.
See, for example, this MO question. The result generalizes to an algebraically closed base field of characteristic zero and this can probably be weakened. For compact groups the above generalizes as follows.
Theorem: Let $G$ be a compact group and $V$ a faithful complex representation of $G$ (so $G$ is a Lie group). Then every irreducible representation of $G$ occurs as a subrepresentation of $V^{\otimes n} \otimes V^{\ast \otimes m}$ for some $n, m$.
See, for example, this MO question. Note also that if the image of $G$ lies in $\text{SL}(V)$, then $\Lambda^{\dim V - 1}(V) \cong V^{\ast}$ (exercise), so $V^{\ast}$ is contained in a tensor power of $V$ and we get the same result as before, with no need to take the dual.