Suppose $U$, $V$, and $W$ are independent and have a standard normal distribution. Then by definition of the chi-squared distribution (as a sum of squares of iid normal variables), the sum of squares $U^2 + V^2 + W^2$ has a chi-squared distribution with three degrees of freedom. We find (via tables or calculation) that its upper 95th percentile is $7.815$. This means that 95% of all the probability lies within the set $$R_{0.95}=\{(u,v,w)\vert u^2 +v^2+w^2 \le 7.815\}.$$
This is a ball of radius $r = \sqrt{7.815}=2.795$; accordingly we obtain its volume $V = 4\pi r^3/3 = 91.51.$
You believe your data are obtained by rescaling $U$ to $X'=e_1U$, $V$ to $Y'=e_2V$, and $Z'=e_3W$ where $e_1,$ $e_2,$ and $e_3$ are the PCA eigenvalues (not their squares!) and then rotating $(X',Y',Z')$ into $(X,Y,Z)$. The rotation does not change volumes, but the initial rescaling means $R_{0.95}$ corresponds to the set
$$\{(x,y,z)\vert (x/e_1)^2 + (y/e_2)^2 + (z/e_3)^2 \le 7.815\}$$
before the rotation occurs. Up to rotation, this is the so-called "confidence ellipse" and therefore has the same shape and volume. Because the orthogonal axes have been separately scaled by $e_1,$ $e_2,$ and $e_3$ (a linear transformation with determinant $e_1e_2e_3$), the new volume is $e_1 e_2 e_3 V = 91.51 e_1 e_2 e_3.$ Similar calculations handle any other level of "confidence," with $7.815$ (and thence $V$) replaced by the associated percentile of a $\chi^2_{(3)}$ distribution.
It may be worth pointing out that a trajectory of a continuously moving point rarely is approximated by a trivariate normal distribution, so this whole program has to be considered approximate and exploratory. As such you could just adopt some standard convenient value for $V$ ($100$ comes to mind, corresponding to $95.96$% of the probability) if your purpose is to compare spreads of trajectories.
