I know that
$$ \cos \theta = \frac{u\cdot e_3}{|u|} $$
but since I neither have any coordinates or the length of $u$, I don't think I can use this. Using the same formula for $u$ on $e_1$ and $e_2$ doesn't help me either. When I first saw it I thougt it'd be simple, just like in $\mathbb{R^2}$, but I can't seem to figure it out.
Since $e_{1}$, $e_{2}$, $e_{3}$ are a basis for $\mathbb{R}^{3}$ we may write:
$$u = a e_{1} + b e_{2} + c e_{3}$$
for some $a, b, c$ in the reals. To find these coefficients, note the property $ e_{i} \cdot e_{j}$ equals $1$ if $i=j$ and $0$ otherwise. Multiply the above equation by $e_{1}$ to get: $a_{1}=u \cdot e_{1}$
Generalizing to find all coefficients gives:
\begin{align} u &= (u \cdot e_{1}) e_{1} + (u \cdot e_{2}) e_{2} + (u \cdot e_{3}) e_{3} \\ &= \| u \| \big( \frac{1}{\sqrt{2}} e_{1} + \frac{1}{2} e_{2} + \cos(x) e_{3}\big) \end{align}
Where $x$ is the angle we're after. Taking the square norm on both sides we get:
$$ \|u\|^2 = \| u \|^2 \big(\frac{1}{2} + \frac{1}{4} + \cos(x)^2\big)$$
Thus $1 = \frac{3}{4} + \cos(x)^2 $, $x = \frac{1}{3}\pi$ or $\frac{5}{3}\pi $
I believe (intuitively) you can also assume WLOG that the basis $\left \{ e_{1}, e_{2}, e_{3} \right \}$ is the standard basis, and that $u$ has unit length, but that might require a more lengthy justification than the above.