Not marking as the answer because I am not fully sure that this is the correct answer to this question.
This is what I came up with after consulting with friends, it does check out, but might not be the right way to solve it:
Theorem: If for some $m=1.1n\log p$ and $n^2<p<2n^2$ is a prime, it holds that for any $k\in\{0,1,...,m\}$:$$\Pr_{e\sim\chi^{\star k}}\left[|e|<\frac{\left\lfloor\frac{p}{2}\right\rfloor}{2}\right] > 1-\delta(n)$$for some negligible function in $n$: $\delta(n)$.
Proof: We first want to clarify the following:$$\text{Sampling }\chi^{\star k}=\text{Sampling }\overline{\Psi}_{\alpha}^{\star k}$$
The latter half is sampling the additive sum distribution of the discretization of $\Psi_{\alpha}$ for $k$. Since $\Psi_{\alpha}$ is the normal distribution (with defined variance and mean according to the overview section) in $\mathbb{T}$, it means that every value sampled from $\Psi_{\alpha}$ will be in $[0,1)$. This means that multiplying it by $p$, will yield a value that is distributed in accordance with the $\Psi_{\alpha}$ distribution but scaled by a factor of $p$.
Due to the above, we can deduce that sampling a single value from $\overline{\Psi}_{\alpha}$ is identical to sampling a value from $\Psi_{\alpha}$, scaling it by $p$ and getting the whole part of it:$$x\sim\Psi_{\alpha}\rightarrow \lfloor px\rceil\sim\overline{\Psi}_{\alpha}$$
Furthering this idea, we can safely say that sampling from the additive sum distribution $\overline{\Psi}_{\alpha}^{\star k}$ is equal to sampling $k$ values from $\overline{\Psi}_{\alpha}$, scaling each by a factor of $p$, rounding and then performing modulo $p$ on the resulting value:$$\begin{array}{c}x_1\sim\overline{\Psi}_{\alpha}, x_2\sim\overline{\Psi}_{\alpha},...,x_k\sim\overline{\Psi}_{\alpha}\\\\
x=\sum_{i=1}^{k}\lfloor px_i\rceil\bmod{p}\\\\
x\sim\overline{\Psi}_{\alpha}^{\star k}\end{array}$$
Given normal probability distributions we know that $\Pr[X<x]=\Phi(z)$ where $z=\frac{x-\mu}{\sigma}$ and $\Phi$ is the CDF of the normal distribution. We will show that for $x=\frac{p}{4}$, the resulting value is $1-\delta(n)$ for a negligible function $\delta(n)$. We first prove that the standard deviation is less than $\frac{p}{\sqrt{\log n}}$ and then we show that the function $1-\Phi(z)$ with the aforementioned $z$ is negligible, and as a result we will see that the above probability is $1-\delta(n)$ for some negligible function $\delta(n)$.
Standard Deviation Upper Bound - we first want to prove that the standard deviation of $\overline{\Psi}^{\star k}_\alpha$ is less than $\frac{p}{\sqrt{\log n}}$. First, due to the properties of standard deviation, and the discretization of $\Psi_{\alpha}$ the standard deviation of $\overline{\Psi}_{\alpha}$ is $\alpha\cdot p$, thus $\text{Var}(\overline{\Psi}_{\alpha})=\alpha^2\cdot p^2$. Due to probability rules and the explanation provided when discussing additive sum distribution, the variance of the additive sum distribution $\overline{\Psi}^{\star k}_\alpha$ is $\text{Var}(\overline{\Psi}^{\star k}_\alpha)=k\cdot\alpha^2\cdot p^2$, which means that the standard deviation of it is $\sqrt{k}\cdot p\cdot\alpha$. The following holds:$$\begin{array}{c}m=1.1\cdot n\log p<1.1\cdot n\log (2n^2)=\\=2.2\cdot n\cdot(\log 2 + \log n)\underset{n>2}{<}4.4n\cdot\log n\end{array}$$
We can therefore say that $m=o(n\log(n))$, therefore we get:$$\sqrt{k}\cdot\alpha\cdot p<\sqrt{m}\cdot\alpha\cdot p=o(\sqrt{n\log n})\cdot o\left(\frac{1}{\sqrt{n}\log(n)}\right)\cdot p =o\left(\frac{1}{\sqrt{\log n}}\right)\cdot p$$
For simplicity we will assume: $\sqrt{k}\cdot\alpha\cdot p = \frac{p}{\sqrt{\log n}}$
Therefore we can say that the standard deviation of $e\sim\overline{\Psi}^{\star k}_{\alpha}$ is less than $\frac{p}{\sqrt{\log n}}$
Negligibility - We know that $\frac{\left\lfloor\frac{p}{2}\right\rfloor}{2}<\frac{p}{4}$ and specifically the difference is less than 1 at most. We also know that - in normal distributions. $$\Pr\left[X<x\right] =\Phi(z), z=\frac{x-\mu}{\sigma}$$ Where $\sigma$ is the standard deviation. For our distribution we know that $\mu=0$ and that $\sigma<\frac{p}{\sqrt{\log{n}}}$ which means that $z=\frac{x}{\sigma}>\frac{x}{\frac{p}{\sqrt{\log{n}}}}$. We define $x=\frac{p}{4}$ which is sufficient in our purposes (as $e$ is taken from $\mathbb{Z}_{p}$ which will only hold whole values, therefore taking $\frac{p}{4}$ gives us a slightly better assurance). We therefore get:$$\Pr_{e\sim\overline{\Psi}^{\star k}_\alpha}\left[|e|<\frac{p}{4}\right]=\Phi\left(\frac{\frac{p}{4}}{\sigma}\right)>\Phi\left(\frac{\frac{p}{4}}{\frac{p}{\sqrt{\log{n}}}}\right)=\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)$$
We also know that $\Phi(z)=1-(1-\Phi(z))$, we will now prove that when $z=\frac{\sqrt{\log{n}}}{4}$, the function $1-\Phi(z)$ is negligible in $n$.
We know from probability theory that $$\begin{array}{c}1-\Phi(z)=1-\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)=1-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\frac{\sqrt{\log{n}}}{4}}e^{-t^2/2}\text{dt}\approx\\\\\approx\frac{1}{\sqrt{2\pi}}\cdot \frac{e^{-\frac{\left(\frac{\sqrt{\log{n}}}{4}\right)^2}{2}}}{\frac{\sqrt{\log{n}}}{4}}=\frac{4}{\sqrt{2\pi\log{n}}}\cdot e^{-\frac{\log{n}}{32}}\end{array}$$
We used the approximation that $1-\Phi(z)\approx\frac{1}{\sqrt{2\pi}\cdot z}e^{-z^2/2}$.
To prove that this is negligilble with regards to $n$, we will show that the following holds $\lim_{n\rightarrow\infty}{(1-\Phi(z))\cdot n^c}=0,\ \ \forall c>0$, as follows:
$$\begin{array}{c}
\lim_{n\rightarrow\infty}{(1-\Phi(z))\cdot n^c}=\lim_{n\rightarrow\infty}{\frac{4}{\sqrt{2\pi\log{n}}}\cdot e^{-\frac{\log{n}}{32}}\cdot n^c}\underset{n^c=e^{c\log n}}{=}\\
=\lim_{n\rightarrow\infty}{\frac{4}{\sqrt{2\pi\log{n}}}\cdot e^{-\frac{\log{n}}{32}+c\log{n}}}=\lim_{n\rightarrow\infty}{\frac{4}{\sqrt{2\pi\log{n}}\cdot{e^{\frac{\log{n}}{32}-c\log{n}}}}}=0
\end{array}$$
As a result, we found that $1-\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)$ is a negligible function.
Therefore we have shown that - given our selection of parameters:
$$\Pr_{e\sim\overline{\Psi}^{\star k}_\alpha}\left[|e|<\frac{p}{4}\right]>\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)=1-\left(1-\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)\right)=1-\delta(n)$$
Where $\delta(n)=1-\Phi\left(\frac{\sqrt{\log{n}}}{4}\right)$ is a negligible function.
