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For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. (The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward.)

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=k[(\mu/\sigma)^2+1]$$ and variance $$V=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward, $k(\mu/\sigma)^2\mapsto\sum_{i=1}^k(\mu_i/\sigma_i)^2={\cal O}(k).$

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=k[(\mu/\sigma)^2+1]$$ and variance $$V=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. (The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward.)

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=\mathbb{E}[Y^2]=k[(\mu/\sigma)^2+1]$$ and variance $$V=\mathbb{E}[Y^4]-\mathbb{E}[Y^2]^2=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. (The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward.)

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=k[(\mu/\sigma)^2+1]$$ and variance $$V=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. (The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward.)

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=\mathbb{E}[Y^2]=k[(\mu/\sigma)^2+1]$$ and variance $$V=\mathbb{E}[Y^4]-\mathbb{E}[Y^2]^2=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. (The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward.)

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=\mathbb{E}[Y^2]=k[(\mu/\sigma)^2+1]$$ and variance $$V=\mathbb{E}[Y^4]-\mathbb{E}[Y^2]^2=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.