I think you should be careful here, as I think you may be confusing a few things. First, the ensemble average used to define the velocity moments are generalized as:
$$
\langle g\left( \mathbf{x}, \mathbf{v} \right) \rangle = \frac{ 1 }{ N } \int d^{3}x \ d^{3}v \ g\left( \mathbf{x}, \mathbf{v} \right) \ f\left( \mathbf{x}, \mathbf{v}, t \right) \tag{0}
$$
where $g(\mathbf{x},\mathbf{v})$ is some dynamical function and $\langle Q \rangle$ is the ensemble average of quantity $Q$. Examples of velocity moments can be found at https://physics.stackexchange.com/a/218643/59023.
An isotropic particle velocity distribution function (VDF) is one where the thermal speeds for each direction are uniform, not the drift speed. That is, the effective width of the distribution function is uniform in all three velocity directions. This says absolutely nothing about whether the distribution is centered on the origin.
The term $\langle v_{i}^{2} \rangle$ is the second velocity moment and technically it should be written as a dyadic product, since the second velocity moment produces the pressure tensor, which is a second rank tensor. Needless to say, the elements of the pressure tensor give you the different thermal speeds along the coordinate basis axes of interest. In most cases, the pressure tensor can be diagonalized to give something like:
$$
\mathbb{P} = \begin{bmatrix}
P_{x} & 0 & 0 \\
0 & P_{y} & 0 \\
0 & 0 & P_{z}
\end{bmatrix} \tag{1}
$$
For an ideal gas, the pressure can be related to the thermal speed (i.e., most probable speed for a 1D Gaussian) as
$$
P_{j} = n \ k_{B} \ T_{j} = \frac{ 1 }{ 2 } n \ m \ V_{Tj}^{2} \tag{2}
$$
where $n$ is the number density, $m$ is the mass, $k_{B}$ is the Boltzmann constant, $T_{j}$ is the jth component of the temperature, and $V_{Tj}$ is the jth component of the thermal speed. Note that the total temperature and/or thermal speed does not sum like a vector but rather they are given by:
$$
\begin{align}
T_{tot} & = \frac{ 1 }{ 3 } \sum_{j} \ T_{j} \tag{3a} \\
V_{T,tot} & = \sqrt{ \frac{ 1 }{ 3 } \sum_{j} \ V_{Tj}^{2} } \tag{3b}
\end{align}
$$
For an isotropic VDF, the pressure tensor elements are uniform, i.e., $P_{x} = P_{y} = P_{z}$, which makes the pressure a scalar. Thus, the corresponding thermal speeds are also uniform which is why $\langle v_{i}^{2} \rangle = \langle \tfrac{v^{2}}{3} \rangle$, i.e., each temperature component is one-third the total.
However, I am not sure how to do this for the average absolute value of a component of the velocity, other than to note that it should be non-zero and should be less than $\langle v \rangle$ itself.
The speed distribution is different than the velocity distribution. For instance, a 3D velocity distribution would look like:
$$
f\left( \mathbf{v} \right) = \frac{ 1 }{ \pi^{3/2} \ V_{T x} \ V_{T y} \ V_{T z} } \ e^{\left[ - \left( \frac{ v_{x} - v_{o, x} }{ V_{T x} } \right)^{2} - \left( \frac{ v_{y} - v_{o, y} }{ V_{T y} } \right)^{2} - \left( \frac{ v_{z} - v_{o, z} }{ V_{T z} } \right)^{2} \right]} \tag{4}
$$
whereas the speed distribution looks like:
$$
\tilde{f}\left( v \right) = \frac{ 4 \ \pi \ v^{2} }{ \pi^{3/2} \ V_{T x} \ V_{T y} \ V_{T z} } \ e^{\left[ - \left( \frac{ v_{x} - v_{o, x} }{ V_{T x} } \right)^{2} - \left( \frac{ v_{y} - v_{o, y} }{ V_{T y} } \right)^{2} - \left( \frac{ v_{z} - v_{o, z} }{ V_{T z} } \right)^{2} \right]} \tag{5}
$$
where $v^{2} = \sum_{j} \ v_{j}^{2}$, $v_{o, j}$ is the displacement of the distribution peak along the jth direction, and $V_{T j}^{2} = \tfrac{ 2 \ k_{B} \ T_{j} }{ m }$ is the most probable speed.
Therefore, to find the any of the most probable speeds, which are the first speed moments in contrast to the thermal speeds resulting from the second velocity moments above, one has:
$$
\langle \lvert v_{i} \rvert \rangle = \int_{0}^{\infty} \ dv \ \lvert v_{i} \rvert \ \tilde{f}\left( v \right) \tag{6}
$$
However, we need to do a change of variables where $dv \rightarrow \tfrac{ v_{i} \ dv_{i} }{ \sqrt{ \sum_{i} \ v_{i} } }$, which makes the integral form go to:
$$
\langle \lvert v_{i} \rvert \rangle = \frac{ 4 \ \pi }{ \pi^{3/2} \ V_{T x} \ V_{T y} \ V_{T z} } \int_{0}^{\infty} \ dv_{i} \lvert v_{i} \rvert^{2} \ \sqrt{ \sum_{j} \ v_{j}^{2} } \ e^{\left[ - \left( \frac{ v_{x} - v_{o, x} }{ V_{T x} } \right)^{2} - \left( \frac{ v_{y} - v_{o, y} }{ V_{T y} } \right)^{2} - \left( \frac{ v_{z} - v_{o, z} }{ V_{T z} } \right)^{2} \right]} \tag{7}
$$
This approach gets messy very fast. The above integral results in a modified Bessel function of the second kind. Note that integrating Equation 5 over all $v$ will give something like $\langle v \rangle = \sqrt{\tfrac{4}{\pi}} \ V_{T,tot}$, which is called the mean speed of the distribution. If the distribution is truly isotropic, then the mean speed should not depend upon the component speed as they are indistinguishable, i.e., $v_{x} \leftrightarrow v_{y} \leftrightarrow v_{z}$, unless the peak of the distribution is not at the origin in all three components in velocity space.
