Partial answer
I thought of a way to transform the problem into a double integral. I don't prove every step, so I can't say I'm a 100% sure this is right. I'm pretty confident that this approach works, but let me know if I made a mistake.
I'm gonna add cosines together instead of sines. It's the same thing, but cosine is a little easier to work with because it's an even function.
Let $n \ge 2$ be the number of cosine functions we're adding together and let $\tau$ be an $n$-dimensional vector containing the rationally independent positive coefficients. We define:
$$
\begin{align}
C &= [-\pi, \pi]^n && \text{($n$ dimensional hypercube)} \\
S &= \left\{x \in C\ \middle|\ \sum_{i=1}^n \cos(x_i) = 0\right\} && \text{($n{-}1$ dimensional surface}) \\
g(x) &= \sum_{i=1}^n \cos(\tau_i x) \\
l_i(x) &= ((\tau_i x + \pi) \operatorname{mod} 2\pi) - \pi && \text{(line through $C$)}
\end{align}
$$
For $n=2$ and $n=3$, $S$ looks like this:
The function $l(x)$ is a line that starts at the origin and goes in direction $\tau$. Whenever it reaches an edge of $C$, it comes out at the edge on the other side.
Now $g(x) = 0$ whenever $l(x) \in S$. So to count the zeroes we can follow the line $l(x)$ and see how often it crosses the surface $S$.
Because of the rational independence, it seems intuitive that the line will travel through each part of $C$ equally often. Therefore we can integrate over the surface $S$ to calculate how often $S$ is crossed.
I figured out the following formula for calculating the frequency $f$. The average distance between zeroes is $1/f$. The function $p(x)$ gives one of the two possible unit normal vectors on the surface $S$ at $x$. The dot represents the dot product of two vectors.
$$
f = \frac{1}{(2\pi)^n} \int_S |p(x) \cdot\tau |\ \mathrm{d}x \label{surfaceint}\tag{1}
$$
This gives for $n=2$:
$$
\begin{align}
f_2 &= \frac{1}{(2\pi)^2} \cdot 2 \pi \sqrt{2} \cdot
\left(\left|
\left[\begin{smallmatrix}\frac12 \sqrt{2} \\ \frac12 \sqrt{2}\end{smallmatrix}\right] \cdot \tau
\right| +
\left|
\left[\begin{smallmatrix}\frac12 \sqrt{2} \\ -\frac12 \sqrt{2}\end{smallmatrix}\right] \cdot \tau
\right|
\right) \\
&= \frac{1}{(2\pi)^2} \cdot 2 \pi \cdot (|\tau_1 + \tau_2| + |\tau_1 - \tau_2|) \\
&= \max(\tau_1, \tau_2) / \pi
\end{align}
$$
For higher $n$, the surface $S$ is more complex and this isn't as easy. Because $S$ is mirror symmetric, we can make it ourselves easier by only integrating over the positive part of $S$. But we do have to take the different normals into account.
$$
\begin{align}
R &= \{x \in S\ |\ \forall_i\ x_i \ge 0\} \\
I(x) &= \sum_{d \in \{-1, 1\}^n} \left| \sum_{i=1}^n d_i \cdot p_i(x) \cdot \tau_i \right| \\
f &= \frac{1}{(2\pi)^n} \int_R I(x)\ \mathrm{d}x
\end{align}
$$
For the normal vector $p(x)$ we can use the normalized gradient of $\sum_{i=1}^n \cos(x_i)$.
I multiplied the whole thing by $-1$ to get a positive normal.
$$
p_i(x) = \sin(x_i) / \sqrt{\sum_{j=1}^n \sin(x_j)^2}
$$
The case $n=3$
Let $u$ be a vector of 3 non-negative elements. The following equation holds:
$$
\sum_{d \in \{-1, 1\}^3} |d \cdot u| = 4 \max (2u_1, 2u_2, 2u_3, u_1+u_2+u_3)
$$
If we combine that equation with the equation for the normal, we get:
$$
I(x) = \frac{
4 \max \left(
\begin{array}{}
2\sin(x_1) \tau_1, \\
2\sin(x_2) \tau_2, \\
2\sin(x_3) \tau_3, \\
\sin(x_1)\tau_1+\sin(x_2) \tau_2+ \sin(x_3) \tau_3
\end{array}
\right)
}
{
\sqrt{\sin(x_1)^2 + \sin(x_2)^2 + \sin(x_3)^2}
}
$$
To rewrite the equation to a normal two dimensional integral, instead of a surface integral,
we first replace $x_3$.
$$
\begin{align}
x_3 &= \arccos(-\cos(x_1)-\cos(x_2)) \\
\sin(x_3) &= \sqrt{1 - (\cos(x_1)+\cos(x_2))^2} \\
I(x) &= \frac{
4 \max \left(
\begin{array}{}
2\sin(x_1) \tau_1, \\
2\sin(x_2) \tau_2, \\
2 \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} , \\
\sin(x_1)\tau_1+\sin(x_2) \tau_2+ \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2}
\end{array}
\right)
}
{
\sqrt{\sin(x_1)^2 + \sin(x_2)^2 + 1 - (\cos(x_1)+\cos(x_2))^2}
}
\end{align}
$$
Now we use the equation:
$$
\begin{align}
\int_R I(x)\ \mathrm{d}x = \int_A I(x)J(x)\ \mathrm{d}x
\end{align}
$$
Where:
$$
\begin{align}
A &= \left\{ \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}x\ \middle|\ x \in R \right\} \\
J(x) &= \sqrt{\left(\frac{\partial x_3}{\partial x_1}\right)^2 + \left(\frac{\partial x_3}{\partial x_2}\right)^2 + 1} \\
&= \sqrt{\frac{\sin(x_1)^2+\sin(x_2)^2+1-(\cos(x_1)+\cos(x_2))^2}{1-(\cos(x_1)+\cos(x_2))^2}}
\end{align}
$$
Combining $I$ and $J$, we get:
$$
I(x)J(x) = \frac{
4 \max \left(
\begin{array}{}
2\sin(x_1) \tau_1, \\
2\sin(x_2) \tau_2, \\
2 \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} , \\
\sin(x_1)\tau_1+\sin(x_2) \tau_2+ \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2}
\end{array}
\right)
}
{
\sqrt{1-(\cos(x_1)+\cos(x_2))^2}
}
$$
So our new integral becomes:
$$
\begin{align}
f_3 &= \frac{1}{(2\pi)^3} \int_A I(x)J(x)\ \mathrm{d}x \\
&= \frac{1}{(2\pi)^3} \left(\int_0^{\frac12\pi} \int_{\arccos(1-\cos(x_1))}^\pi I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1 +
\int_{\frac12\pi}^\pi \int_0^{\arccos(-1-\cos(x_1))} I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1 \right) \\
&= \frac{1}{4\pi^3} \int_0^{\frac12\pi} \int_{\arccos(1-\cos(x_1))}^\pi I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1
\end{align}
$$
We can get rid of those nasty sines and cosines by using integration by substitution.
Replacing $x_2$ with $\arccos(v)$ gives:
$$
\begin{align}
H(x_1, v) &= \frac{
\max \left(
\begin{array}{}
2\sin(x_1) \tau_1, \\
2\tau_2\sqrt{1-v^2}, \\
2 \tau_3\sqrt{1 - (\cos(x_1)+v)^2} , \\
\sin(x_1)\tau_1+\tau_2\sqrt{1-v^2} + \tau_3\sqrt{1 - (\cos(x_1)+v)^2}
\end{array}
\right)
}
{
\sqrt{1-v^2} \cdot \sqrt{1-(\cos(x_1)+v)^2}
} \\
f_3 &= \frac{1}{\pi^3} \int_0^{\frac12\pi} \int_{-1}^{1-\cos(x_1)} H(x_1, v) \ \mathrm{d}v\ \mathrm{d}x_1
\end{align}
$$
Then replacing $x_1$ with $\arccos(u)$ gives:
$$
\begin{align}
G(u, v) &= \frac{
\max \left(
\begin{array}{}
2\tau_1\sqrt{1-u^2} , \\
2\tau_2\sqrt{1-v^2}, \\
2 \tau_3\sqrt{1 - (u+v)^2} , \\
\tau_1\sqrt{1-u^2}+\tau_2\sqrt{1-v^2} + \tau_3\sqrt{1 - (u+v)^2}
\end{array}
\right)
}
{
\sqrt{1-u^2} \cdot \sqrt{1-v^2} \cdot \sqrt{1-(u+v)^2}
} \\
f_3 &= \frac{1}{\pi^3} \int_0^1 \int_{-1}^{1-u} G(u, v) \ \mathrm{d}v\ \mathrm{d}u
\end{align}
$$
One possible way to continue solving this integral, is splitting it into four integrals,
one for each of the arguments of $\max$. To do this we need to find the values for
$u$ and $v$ in which these arguments are the maximum. The conditions can be reduced to:
$$
\begin{align}
1\colon&&\ \tau_1 \sqrt{1-u^2} &> \tau_2 \sqrt{1-v^2} + \tau_3 \sqrt{1-(u+v)^2} \\
2\colon&&\ \tau_2 \sqrt{1-v^2} &> \tau_1 \sqrt{1-u^2} + \tau_3 \sqrt{1-(u+v)^2} \\
3\colon&&\ \tau_3 \sqrt{1-(u+v)^2} &> \tau_1 \sqrt{1-u^2} + \tau_2 \sqrt{1-v^2} \\
4\colon&&\ \text{otherwise}
\end{align}
$$
Numerical approximation
The following Mathematica code calculates an approximation to $\ref{surfaceint}$.
I tried to write it to work in any number of dimensions, but it only seems
to work in exactly 3 dimensions (Mathematica 11.2).
frequency[t_] := Module[{n, vars, x, r},
n = Length[t];
vars = Table[x[i], {i, n}];
r = ImplicitRegion[Total[Cos[vars]] == 0, Evaluate[{#, -Pi, Pi}& /@ vars]];
NIntegrate[Abs[t.Normalize[Sin[vars]]], vars \[Element] r] / (2Pi)^n
];
Print["Average distance between zeroes: ", 1 / frequency[{1, Sqrt[2], Sqrt[3]}]];
The code outputs $2.22465$. I don't know how many digits of that are right.
