Why are randomly drawn vectors nearly perpendicular in high dimensions

A random uniform unit vector is $X/\|X\|$ where $X$ is standard normal, thus the scalar product of two independent unit vectors $U$ and $V$ is $\langle U,V\rangle=\langle X,Y\rangle/(\|X\|\cdot\|Y\|)$ where $X$ and $Y$ are independent and standard normal. When $n\to\infty$, by the law of large numbers, $\|X\|/\sqrt{n}\to1$ almost surely and $\|Y\|/\sqrt{n}\to1$ almost surely, and by the central limit theorem, $\langle X,Y\rangle/\sqrt{n}$ converges in distribution to a standard one-dimensional normal random variable $Z$.

Thus, $\sqrt{n}\cdot\langle U,V\rangle\to Z$ in distribution, in particular, for every $\varepsilon\gt0$, $P(|\langle U,V\rangle|\geqslant\varepsilon)\to0$. In this sense, when $n\to\infty$, the probability that $U$ and $V$ are nearly orthogonal goes to $1$.

Likewise, $k$ independent uniform unit vectors are nearly orthogonal with very high probability when $n\to\infty$, for every fixed $k$.

Here is one way to reason, chosen for simplicity of calculations: Consider the unit vector $e=(1,0,0,\ldots,0)\in\mathbb R^n$. One way to measure how 'orthogonal' $e$ is to other vectors is to calculate the average of $(e\cdot x)^2$ as $x$ ranges over the unit sphere. If $S$ denotes the surface measure on the unit sphere corresponding to (normalized) area, then $$ \int |e\cdot y|^2 dS(y) =\int |y_1|^2 dS(y)=\frac{1}{n}\int \sum_{j=1}^n |y_j|^2 dS(y)=\frac{1}{n}. $$ Thus, in this sense, vectors are generally 'more' orthogonal in higher dimensional spaces.

Edit: This line of reasoning follows closely the argument given by JyrkiLahtonen in the comments above, as one sees by considering a random $\mathbb R^n$-valued vector $Y$, uniformly distributed on the unit sphere. If we consider the random variable $e\cdot Y$, then $$ E \; e\cdot Y=\int e\cdot y \;dS(y)=0, $$ because $S$ in invariant under the transformation $y\mapsto -y$. On the other hand $$ V(e\cdot Y)=\int |e\cdot y|^2 dS(y) =\frac{1}{n}, $$ as shown above. Therefore, intuitively, $e\cdot Y$ is small when $n$ is large. Rigorously, we can employ Chebyshev's inequality to obtain $$ P(|e\cdot Y|\geq \epsilon)\leq \frac{1}{n\epsilon^2}. $$

Here's an excerpt from Lecture 2 of Keith Ball's An Elementary Introduction to Modern Convex Geometry:

See the link for details, but as the image shows, the measure of a spherical cap cut off by a hyperplane which is $\varepsilon$ away from the origin is bounded above by the ratio between volume of a ball of radius $\sqrt{1-\varepsilon^2}$ and the volume of a ball of radius 1, that is, $(1-\varepsilon^2)^{n/2}$. So if $X$ and $Y$ are independent random unit vectors (uniform on the sphere), then $$ P(\langle X,Y\rangle < \varepsilon) \ge 1 - (1-\varepsilon^2)^{n/2} \ge 1 - e^{-n\varepsilon^2/2} $$ which is close to $1$ when $n$ is large (if $\varepsilon$ is fixed, at least).