Let me separate the radial integration from the angular integration,
$$\int_{|\mathbf{x}|\leq 1}f(\mathbf{x})d\mathbf{x}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_0^1 r^{n-1}\bar{f}(r)\,dr,$$
where $\bar{f}(r)$ is the average of $f$ over the surface of the $n$-dimensional hypersphere of radius $r$. In our case
$$f(\mathbf{r})=\bigg(\sum_{j=1}^{n}a_{j}x_{j}^2\bigg)^m.$$
For $n\gg 1$ the concentration of measure allows us to replace $x_j^2$ by $r^2/n$, so 
$$\bar{f}(r)\approx (r^2/n)^m\bigg(\sum_{j=1}^{n}a_{j}\bigg)^m,$$
and thus we estimate
$$\int_{B(0,1)}\bigg(\sum_{j=1}^{n}a_{j}x_{j}^2\bigg)^m \,d\mathbf{x}\approx
\frac{2\pi^{n/2}}{n^m(2m+n)\Gamma(n/2)}\bigg(\sum_{j=1}^{n}a_{j}\bigg)^m,\;\;n\gg 1.$$

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Obviously, this estimate is exact if all $a_j$'s are equal to each other. To check for the other extreme, let me try (following the suggestion by Iosif Pinelis) the simple case $m=1$, $a_1=1$ and $a_j=0$ for $j=2,3,\ldots n$. The exact integration gives
$$\int_{B(0,1)}x_1^2\, d\mathbf{x}=\frac{2 \pi ^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int_0^1 r^{n+1}dr\int_0^\pi\cos^2\phi\sin^{n-2}\phi d\phi=  \frac{2\pi^{n/2}}{n(n+2)\Gamma(n/2)},$$
which again equals the large-$n$ estimate.