I'm assuming the vectorized $\eta$ is uniformly distributed on the sphere of radius $1/10$ in $\mathbb R^{n^2}$.

For fixed $x$, $$ \mathbb E \| \eta x \|_2^2 = \mathbb E \sum_i \sum_j \sum_k \eta_{ij} \eta_{ik} x_j x_k = \sum_j \sum_k \mathbb E (\eta^T \eta)_{jk} x_j x_k $$

Now for $j = k$, $$ \mathbb E(\eta^T \eta)_{kk} = \sum_i \mathbb E \eta_{ik}^2 = \frac{1}{n} \sum_{i}\sum_j \eta_{ik}^2 = \frac{1}{100 n}$$ while for $j \ne k$, since the conditional distribution of $\eta_{ij}$ given $\eta_{ik}$ is symmetric about $0$, $$\mathbb E(\eta^T \eta)_{jk} = \sum_i \mathbb E(\eta_{ij} \eta_{ik}) = 0$$ Thus $$ \mathbb E \|\eta x\|_2^2 = \sum_k \frac{1}{100 n}x_k^2 = \frac{\|x\|_2^2}{100n}$$

I'm assuming the vectorized $\eta$ is uniformly distributed on the sphere of radius $1/10$ in $\mathbb R^{n^2}$.

For fixed $x$, $$ \mathbb E \| \eta x \|_2^2 = \mathbb E \sum_i \sum_j \sum_k \eta_{ij} \eta_{ik} x_j x_k = \sum_j \sum_k \mathbb E (\eta^T \eta)_{jk} x_j x_k $$

Now for $j = k$, $$ \mathbb E(\eta^T \eta)_{kk} = \sum_i \mathbb E \eta_{ik}^2 = \frac{1}{n} \sum_{i}\sum_j \eta_{ik}^2 = \frac{1}{100 n}$$ while for $j \ne k$, since the conditional distribution of $\eta_{ij}$ given $\eta_{ik}$ is symmetric about $0$, $$\mathbb E(\eta^T \eta)_{jk} = \sum_i \mathbb E(\eta_{ij} \eta_{ik}) = 0$$ Thus $$ \mathbb E \|\eta x\|_2^2 = \sum_k \frac{1}{100 n}x_k^2 = \frac{\|x\|_2^2}{100n}$$

I suspect more can be said about the distribution of $\|\eta x \|$ using concentration of measures, but I'll leave that to someone else.

For fixed $x$, $$ \mathbb E \| \eta x \|_2^2 = \mathbb E \sum_i \sum_j \sum_k \eta_{ij} \eta_{ik} x_j x_k = \sum_j \sum_k \mathbb E (\eta^T \eta)_{jk} x_j x_k $$

Now for $j = k$, $$ \mathbb E(\eta^T \eta)_{kk} = \sum_i \mathbb E \eta_{ik}^2 = \frac{1}{n} \sum_{i}\sum_j \eta_{ik}^2 = \frac{1}{100 n}$$ while for $j \ne k$, since the conditional distribution of $\eta_{ij}$ given $\eta_{ik}$ is symmetric about $0$, $$\mathbb E(\eta^T \eta)_{jk} = \sum_i \mathbb E(\eta_{ij} \eta_{ik}) = 0$$ Thus $$ \mathbb E \|\eta x\|_2^2 = \sum_k \frac{1}{100 n}x_k^2 = \frac{\|x\|_2^2}{100n}$$

I'm assuming the vectorized $\eta$ is uniformly distributed on the sphere of radius $1/10$ in $\mathbb R^{n^2}$.

For fixed $x$, $$ \mathbb E \| \eta x \|_2^2 = \mathbb E \sum_i \sum_j \sum_k \eta_{ij} \eta_{ik} x_j x_k = \sum_j \sum_k \mathbb E (\eta^T \eta)_{jk} x_j x_k $$

Now for $j = k$, $$ \mathbb E(\eta^T \eta)_{kk} = \sum_i \mathbb E \eta_{ik}^2 = \frac{1}{n} \sum_{i}\sum_j \eta_{ik}^2 = \frac{1}{100 n}$$ while for $j \ne k$, since the conditional distribution of $\eta_{ij}$ given $\eta_{ik}$ is symmetric about $0$, $$\mathbb E(\eta^T \eta)_{jk} = \sum_i \mathbb E(\eta_{ij} \eta_{ik}) = 0$$ Thus $$ \mathbb E \|\eta x\|_2^2 = \sum_k \frac{1}{100 n}x_k^2 = \frac{\|x\|_2^2}{100n}$$