My linked list results in segmentation fault every time even though I cannot see a flaw

In the function push_front,

cell* n;

is a pointer of type cell and as David C Rankin rightly said it is uninitialized and currently holds an indeterminate address as its value until you initialize it by assigning a valid address. It does not point to a memory of type cell just like that until you make it point to a cell type.

Use malloc to allocate memory. You can refer the manual page for malloc.

cell *n= malloc(sizeof(cell));

The very next thing your code should do is to check if call to malloc has successfully allocated memory or not and some error handling if memory is not allocated.

if(NULL == n)
{
    //! handle memory allocation error here
}

You also need to free this memory once you are done using it, and make the pointer point to null.

free(n);
n = NULL;

Based on your critical rant/comment at the end of original post, let me explain the code you wrote to you.

void push_front(cell_list* list, int number)
{
    /// This print helps you understand where the code is.
    printf("here\n");
    /// This declares is a pointer variable pointing to a `cell` type. 
    /// Not yet pointing since not initialized.
    cell* n;
    /// This print helps you understand where the code is.
    printf("here2\n");
    /// Tries to assign value of 'number' to 'number' variable present in 
    /// some memory accessed as data type `cell` via a pointer. 
    /// But wait, is pointer really pointing to a valid memory location 
    /// that can be access as `cell` type? NO!
    n->number = number; // You try to sit in a chair which isn't there, you fall!
    printf("here3\n");
    n->next = list->head; // You try to sit on a chair which isn't there, you fall!
    printf("here4\n");
    list->head = n; // This is a different problem as of now. 
                    // With 'n' not initialized, list->head also get
                    // corrupted since assigned with a garbage value. 
}

Also, in your main() function,

cell_list* L;

and then you invoke

push_front(L, 5);

In this function, you do the following:

n->next = list->head;

You design does not consider if list->head is initialized or not. If it is NULL you just made n->next point to NULL.

I understand and agree to the fact that a lot of people find pointers hard to understand and use correctly. I'll suggest that you first get comfortable with pointers overall (in general) and then go for implementing data structures with them. We all started there! You can start here.

You have to allocate memory for nodes using malloc. Here is the updated code, I have tested it, it is working fine :)

#include <stdio.h>
#include <stdlib.h>

typedef struct cell {
    int number;
    struct cell* next;
} cell;

typedef struct {
    cell* head;
} cell_list;


void push_front(cell_list* list, int number) {
    printf("here\n");
    cell* n = malloc(sizeof(cell));
    printf("here2\n");
    n->number = number;
    printf("here3\n");
    n->next = list->head;
    printf("here4\n");
    list->head = n;
}

int main(int argc, char* argv[]) {
    cell_list* L = malloc(sizeof(cell_list));

    push_front(L, 5);
    push_front(L, 8);
    push_front(L, 19);

    cell* n = L->head;

    // this should print "19 8 5"
    while(n != NULL) {
        printf("Val: %d\n", n->number);
        n = n->next;
    }

    return 0;
}