One pass solution with backtracking, collecting all valid palindromes. Passes all tests.

def substrCount(n, s):
    # print('---', s)
    subs = []
    for i, l in enumerate(s):
        # collect all single chars
        subs.append(l)

        if i > 0:
            # collect all substrings of same chars behind i
            j = i - 1
            sub = l
            while j >= 0 and s[j] == l:
                sub += s[j]
                subs.append(sub)
                j -= 1

            # collect all substrings where middle char is different 
            # and all chars on left and right are the same
            j = i - 1
            if s[j] != l:                
                k = i + 1
                sub = l
                while j >= 0 and k < n and s[j] == s[k]:
                    if j+1 < i and s[j] != s[j+1]:
                        break

                    sub = f'{s[j]}{sub}{s[k]}'
                    subs.append(sub)
                    j -= 1
                    k += 1

    # print(subs)

    return len(subs)