This question was asked at math.stackexchange, where it got several upvotes but no answers.

It is impossible to find $n+1$ mutually orthonormal vectors in $R^n$.

However, it is well established that the central angle between legs of a regular simplex with $n$-dimensional volume goes as $\theta = \mathrm{arccos}(-1/n)$. This approaches $90$ degrees as $n \rightarrow \infty$, so since there are $n+1$ vertices of a simplex with $n$-dimensional volume, we can conclude

Given $\epsilon > 0$, there exists a $n$ such that we can find $n+1~$

approximately mutually orthogonalvectors in $\mathbb{R}^n$, up to tolerance $\epsilon$. (Unit vectors $u$ and $v$ are said to be approximately orthogonal to tolerance $\epsilon$ if their inner product satisfies $\langle u,v \rangle < \epsilon$)

My question is a natural generalization of this - if we can squeeze $n+1$ approximately mutually orthonormal vectors into $\mathbb{R}^n$ for $n$ sufficiently large, how many more vectors can we squeeze in? $n+2$? $n+m$ for any $m$? $2n$? $n^2$? $e^n$?

Actually the $n+m$ case is easy to construct from the $n+1$ case. Given $\epsilon$, one finds the $k$ such that you can have $k+1$ $\epsilon$-approximate mutually orthogonal unit vectors in $\mathbb{R}^k$. Call these vectors $v_1, v_2, ..., v_k$. Then you could squeeze $mk+m$ vectors in $\mathbb{R}^{mk}$, by using the vectors $$\begin{bmatrix} v_1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_2 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_{k+1} \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_1 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_2 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_{k+1} \\ \vdots \\ 0 \end{bmatrix}, \dots \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_1 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_2 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_{k+1} \\ \end{bmatrix}. $$ So, setting $n = mk$, we have found an $n$ such that we can fit $n + m$ $\epsilon$-orthogonal unit vectors in $\mathbb{R}^n$.

I've haven't been able to construct anything stronger than $n+m$, but I also haven't been able to show that this is the upper bound.