How to calculated a probability for a sum of integers to be equal to a given value if probabilities of addends are known? Let's assume that we have a process generating some integer numbers with different probabilities. The set of possible integers is small. For example: -2, -1, 0, 1 and 2. We also know probabilities for each number to be generated.
Now we generate N numbers and sum them up. We want to know the probability of the sum to be equal to, let's say, 37. Is there an analytical solution for this problem?
The closes that I found is multinomial distribution. With this distribution we can calculate the probability that, let's say, -2 was generated n1 times, -1 was generated n2 times and so on. However, I am interested in the sum and there are many different ways to get the considered sum.
 A: It seems there are some gaps in assumptions and in the probabilities assigned to Likert categories. In addition to the suggestions in @whuber's link, here are possible approaches via normal approximation and simulation.
Normal approximation. By the Central Limit Theorem, the sum of $n$ is very nearly normal for moderately large $n.$ Let $Y_i$ be the number on the $i$th die out of $n$ and $X$ be the total on all $n$ dice. Depending on assumptions, it is easy to find $E(Y_i), V(Y_i)$ and hence $E(X)$ and ${}_{}($for independent $Y_i,)$ also $SD(X).$ Finally, you can use a normal approximation, with continuity correction, to find the approximate value of $P(X = 37).$
In particular, suppose each question on a questionnaire has
possible Likert responses $Y$ coded -2 through 2 with respective probabilities in the vector pp=c(.1, .2, .2, .3, .2). Then $E(Y) = 0.3$ and $V(Y) = 1.61.$
If a questionnaire has $n = 70$ independent such questions, then the total score $X$ has $E(X) = 21$ and $SD(X) = 10.616.$
n = 70;  lik = -2:2;  pp = c(.1, .2, .2, .3, .2)
mu = sum(lik * pp);  mu
[1] 0.3                         # E(Y)
mu.x = n*mu;  mu.x
[1] 21                          # E(X)
vr = sum(lik^2 * pp) - mu^2;  vr
[1] 1.61                        # V(Y)  
sd.x = sqrt(n * vr);  sd.x
[1] 10.61603                    # SD(x)

Simulation. Then the score of a randomly chosen subject might
be simulated in R as sample(-2:2, 70, rep=T, prob=pp). If we look at a million simulated scores, then we can (a) find  the proportion of scores that are $37,$
(b) find the median score, and (c) get other information
about the distribution of the scores:
set.seed(2019)
m = 10^6;  n = 70
lik = -2:2;  pp=c(.1, .2, .2, .3, .2)
x = replicate(m, sum(sample(lik, n, rep=T, prob=pp)))
mean(x==37)
[1] 0.012153
median(x)
[1] 21
summary(x)
  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   -38      14      21      21      28      70 

Comparisons. The figure below compares the simulated distribution of $X$ with the density function of $\mathsf{Norm}(21, 10.616).$

From normal approximation, $P(X = 37) \approx 0.0121,$
which in good agreement with the simulated value. 
diff(pnorm(c(36.5,37.5), 21, 10.61603))
[1] 0.01207532

