Assume the "summary effect size" is the inverse variance weighted mean
$$M = \frac{1}{\sum_{i=1}^k (1/V_i)}\sum_{i=1}^k \frac{Y_i}{V_i} = \sum_{i=1}^k \omega_i Y_i$$
(thereby defining the weights $\omega_i$ as a notational convenience). Let $\Omega$ be the denominator
$$\Omega = \sum_{i=1}^k \frac{1}{V_i}$$
so that
$$\omega_i = \frac{1/V_i}{\Omega}.$$
We will need to do some algebra. Observe, for later, that
$$\sum_{i=1}^k \omega_i = 1$$
and
$$\Omega \omega_i V_i = \Omega \frac{1/V_i}{\Omega} V_i = 1.$$
I will use these relations repeatedly without further comment.
One of the least painful approaches begins by expanding the terms of $Q$ as
$$\frac{1}{V_i}(Y_i-M)^2 = \Omega\omega_i\left(Y_i - \sum_{j=1}^k \omega_jY_j\right)^2=\Omega\omega_i\left((1-\omega_i)^2Y_i^2 + \sum_{j\ne i}\omega_j^2 Y_j^2 + \cdots\right)$$
where "$\cdots$" comprises multiples of the cross-products $Y_iY_j$ for $j\ne i.$
Assuming all the $Y_i$ have a common expectation $\mu,$ notice that changing the units of measurement of the $Y_i$ by subtracting $\mu$ from them does not change the foregoing values, because (i) the variances $V_i$ remain the same; and because the weights $\omega_i$ sum to unity, (ii) $M$ changes to
$$M^\prime = \sum_{i=1}^k \omega_i (Y_i - \mu) = \left(\sum_{i=1}^k \omega_i Y_i\right) - \mu.$$
The terms $Y_i-M$ thereby change to $(Y_i-\mu)-(M-\mu) = Y_i-M:$ that is, they are unaltered. Thus, with no loss of generality, we may assume the expectation of each $Y_i$ is zero, whence
$$V_i = \operatorname{Var}(Y_i) = E[Y_i^2]$$
and for $i\ne j,$
$$E[Y_iY_j] = (0)(0) = 0$$
(by independence) . Upon taking the expectation of $Q,$ then, those cross-terms all drop out (which is why we didn't need to compute their coefficients), leaving
$$E[Q] = \sum_{i=1}^k \Omega\omega_i E\left[(1-\omega_i)^2Y_i^2 + \sum_{j\ne i}\omega_j^2 Y_j^2 + \cdots\right] = \sum_{i=1}^k \Omega\omega_i\left((1-\omega_i)^2 V_i + \sum_{j\ne i} \omega_j^2 V_j\right).$$
As written, this has been expressed as a linear combination of the $V_i.$ By inspection, the coefficient of any particular $V_i$ is the sum of the coefficient from the $i^\text{th}$ term plus all the coefficients of $V_i$ that show up in the second sum. The total contribution (including the factor of $V_i$) to the expectation is
$$\Omega \omega_i V_i \left((1-\omega_i)^2 + \omega_i\sum_{j\ne i}\omega_j\right) =(1-\omega_i)^2 + \omega_i(1-\omega_i) = 1-\omega_i. $$
Consequently
$$E[Q] = \sum_{i=1}^k (1-\omega_i) = k - \sum_{i=1}^k \omega_i = k-1,$$
QED.
NB: This proof requires no knowledge of the distribution of $Q$. That's good, because $Q$ rarely has a chi-squared distribution. But when the $Y_i$ are independent Normal variables with common expectation, then $Q$ does turn out to have a chi-squared distribution. (This is not immediately evident, because $Q$ is explicitly the sum of non-independent terms due to the common appearance of $M$ in each numerator. This result is a consequence of Cochran's Theorem.)
