My first inclination is to say this is a meaningless question. The concept of "instantaneous" frequency really only pertains to a single pure tone with a slightly varying frequency.
In this light, one may construct a definition saying "The instantaneous frequency at time $t$ is the same as that of a pure tone which matches the function (sum) in the immediate time neighborhood".
I think you have the signs wrong in your statement, and they are conceptual clutter, so I have removed them. The same solution can be done with them retained.
$$ A e^{j (\omega t + \theta) }=e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)} $$
Take the derivative of both sides.
$$ j \omega A e^{j (\omega t + \theta) }= j \left( \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} \right) \frac{d\phi}{dt} $$
Cancel the $j$s and divide both sides by the single tone definition.
$$ \omega = \frac{ \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} }{A e^{j (\omega t + \theta) }} \cdot \frac{d\phi}{dt} $$
Substitute it back in.
$$ \omega = \frac{ \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} }{e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)}} \cdot \frac{d\phi}{dt} $$
You can now solve for $\omega$. It may be complex. The real part represents the rotational frequency and the imaginary part the decay in amplitude coefficient.
The answer is then $\frac{1}{2\pi}\Re(\omega$), where the fraction is the conversion factor between radians per unit time and cycles per unit time. $2\pi$ has units of radians per cycle.
After thoughts:
I think this answer qualifies under R B-J's "overall" categorization.
The constraint that the $\alpha$'s be integer value seems superfluous. All that does is ensure that the overall signal is periodic. The same would be true for rational $\alpha$ values too. This has nothing to do with local behavior.
Also, if you plug in $\alpha_1=-1,\alpha_2=0,\alpha_3=0$, you can see the OP's original assertion has a sign error, otherwise, they are in agreement.
You might find the formulas I derive in these articles interesting and pertinent.
These formulas are derived based on the assumption of a single pure tone which varies only in frequency. Surprisingly, the same formulas work for both the real and complex tone cases. Obviously, being at a peak or a zero crossing is only relevant in the real tone case.
Since the assumption of my solution is that the sum is acting like a single pure tone in a small neighborhood, the article formulas, based on a few neighboring points, should return close to the same results as the derived solution in my answer, based on the signal definition.
Per LD, the "close to a kind of an average of the base frequencies" formulation:
$$ S = e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)} $$
$$ \omega = \left[ \left( \frac{e^{j \alpha_1 \phi(t)}}{S} \right) \cdot \alpha_1 + \left( \frac{e^{j \alpha_2 \phi(t)}}{S} \right) \cdot \alpha_2 + \left( \frac{e^{j \alpha_3 \phi(t)}}{S} \right)\cdot \alpha_3 \right] \cdot \frac{d\phi}{dt} $$
Instead of matching the signal locally to a complex pure tone, it can instead be matched to a real pure tone.
$$ A \cos( \omega t + \phi ) = S $$
$$ -\omega A \sin( \omega t + \phi ) = S' $$
$$ -\omega^2 A \cos( \omega t + \phi ) = S'' $$
$$ \omega^2 = -\frac{S''}{S} $$
$$ \omega = j\sqrt{\frac{S''}{S}} $$
The reader is invited to plug the definition of $S$ in from above.
If you do plug in the the original $S$ in the real tone matching case, you will get an RMS like formula which is also a kind of weighted average of the base frequencies.
Note too, with these formulas, it is possible for the denominator to become zero while the numerator is non-zero. Thus the instantaneous frequency is supposedly infinite (loosely speaking). That doesn't make sense either.
To see that the two approaches don't necessarily reach the same value it is easier to use a simpler function.
Suppose
$$ S = t^2 $$
The two different approaches yield different results:
$$ \omega_{1} = -j\frac{S'}{S} = \frac{-2}{t}j $$
$$ \omega_{2} = j\sqrt{\frac{2}{t^2}} = \frac{\sqrt{2}}{|t|}j $$
There is nothing that says that either of these methods is any more or less valid than the other.
Taking a totally different tack, let's look at the specific case of
$$ \alpha_1 = 3, \alpha_2 = 7,\alpha_3 = 0, \phi(t)=t $$
$$
\begin{align}
S &= e^{j3t} + e^{j7t} \\
&= e^{j5t} \left( e^{-j2t} + e^{j2t} \right) \\
&= 2 e^{j5t} \cos(2t)
\end{align}
$$
This specific example has an unwavering rotational frequency of 5 radians per unit time and an unwavering amplitude oscillation frequency of 2 radians per unit time. This is the complex analog of the beat phenomenon. Since both are unwavering, the "instantaneous" values should be the same.
On the other hand, the signal repeats itself every $2\pi$ units of time, so its overall global frequency is 1 radian per unit time. It is an example of a periodic signal missing its fundamental.
So, what is the actual frequency?
I return to my original inclination, this is a meaningless question.