Suppose I sample $n$ (some number) numbers from a Normal distribution with zero mean and standard deviation $\sigma$. That is, from $\mathcal{N}(0,\sigma)$. We will call this list of numbers $L$. We will take $L(i)$ to mean the $i$th element of $L$.
I now want to create a function using these numbers. It will be a very simple function. It will be piecewise-constant, with each constant segment having the same length and the constant value corresponding to the values in $L$. A picture of this function for $n=30$ is given underneath,
If I now sampled this function finely (say 10 samples per constant segment), and I made a histogram of the sampled values, I believe that the histogram will reproduce a $\mathcal{N}(0,\sigma)$ distribution.
Now, we consider another function, where the elements of $L$ are instead the vertices of a piecewise-linear function. For $n = 30$, this function looks like 
That is, there is a linear interpolation between each of the vertices, which are the elements of $L$. If now now sampled this function finely, and I make a histogram of the sampled values, would the histogram also be like $\mathcal{N}(0,\sigma)$? Would it be for large $n$?
Let's say we have $$L \sim \mathcal{N}(0,\sigma^2) $$And you have another random discrete variable $S$, which consists of discretization levels. One scheme is as such (PS: I'm just trying to mimic your figure because you did not provide any extra information on the sampling scheme) \begin{align} f(L = l) &= 0 \qquad \text{ if } \vert l \vert < \frac{W}{2} \\ f(L = l) &= A \qquad \text{ if } (1-\frac{1}{2})W< l < (1+\frac{1}{2})W \\ f(L = l) &= -A \qquad \text{ if } - (1+\frac{1}{2})W< l < - (1-\frac{1}{2})W \\ f(L = l) &= 2A \qquad \text{ if } (2-\frac{1}{2})W< l < (2+\frac{1}{2})W \\ f(L = l) &= -2A \qquad \text{ if } - (2+\frac{1}{2})W< l < - (2-\frac{1}{2})W \end{align} So in general the sampling scheme is \begin{equation} f(L = l) = kA, \qquad \text{ if } (k-\frac{1}{2})W< l < (k+\frac{1}{2})W \end{equation} and \begin{equation} f(L = l) = -kA, \qquad \text{ if } -(k+\frac{1}{2})W< l < -(k-\frac{1}{2})W \end{equation} Then your distribution is just \begin{equation} Pr(S = kA) = Pr((k-\frac{1}{2})W< l < (k+\frac{1}{2})W ) \end{equation} which by the definition of the Normal distribution is \begin{equation} Pr(S = kA) = \int\limits_{(k-\frac{1}{2})W}^{(k+\frac{1}{2})W } \Phi_{\sigma}(x) \ dx \qquad k = 0 \ldots (K-1) \end{equation} Similary which by the definition of the Normal distribution is \begin{equation} Pr(S = -kA) = -\int\limits_{-(k-\frac{1}{2})W}^{-(k+\frac{1}{2})W } \Phi_{\sigma}(x) \ dx \qquad k = -(K-1) \ldots -1 \end{equation} where $\Phi_{\sigma}(x)$ is the PDF of a zero mean normal distribution with standard deviation $\sigma$ and $KA$ is the largest sampling level.
NOTE: There is one exception on this sampling distribution, which is the largest sampling level: $$f(L=l) = KA \qquad \text{ if } (K-\frac{1}{2})W< l$$then \begin{equation} Pr(S = KA) = \int\limits_{(K-\frac{1}{2})W}^{+\infty} \Phi_{\sigma}(x) \ dx \end{equation} and \begin{equation} Pr(S = -KA) = \int\limits_{-\infty}^{-(K+\frac{1}{2})W} \Phi_{\sigma}(x) \ dx \end{equation}