This is not a complete answer, but grew large for a comment.

Duality acts on $\{\omega_i\}$, by $-w_0$, so we should really group them into the set of orbit sums $D = \{\omega_i$ self-dual, $\omega_j - w_0\cdot \omega_j$ not$\}$. Then $\lambda = \sum_{d \in D} \lambda_d d$. Let $Q = \{d\in D\ :\ V_d$ is quaternionic$\}$.

(Is it obvious that $Q$ doesn't contain any of the sums $\omega_j - w_0\cdot \omega_j$? I feel that should be true and obvious.)

It's easy to see that the indicator is multiplicative in $\lambda$, by tensoring the forms together, which leads to the formula $(-1)^{\sum_Q \lambda_q}$. (Of course Mikhail Borovoi's answer is of this form.)

This is not a complete answer, but grew large for a comment.

Duality acts on the simple weights $\{\omega_i\}$, by $-w_0$ (where $w_0$ is the long element of the Weyl group), so we should really group them into the set of orbit sums $$ D = \{\omega_i\,|\,\omega_i\,\, \text{is self-dual}\}\cup \{\omega_j - w_0\cdot \omega_j\,|\,\omega_j\,\, \text{is not self-dual}\}. $$ Then $\lambda = \sum_{d \in D} \lambda_d d$. Let $Q = \{d\in D\ :\ V_d$ is quaternionic$\}$.

(Is it obvious that $Q$ doesn't contain any of the sums $\omega_j - w_0\cdot \omega_j$? I feel that should be true and obvious.)

It's easy to see that the indicator is multiplicative in $\lambda$, by tensoring the forms together, which leads to the formula $(-1)^{\sum_Q \lambda_q}$. (Of course Mikhail Borovoi's answer is of this form.)