Let's look at an example with \$N=3\$ stages. (This can be generalized to any \$N\$.)
simulate this circuit – Schematic created using CircuitLab
I'll assume for your purposes that, for all \$N=3\$, \$Z_{\text{A}_i}=Z_{_\text{A}}=\frac1{s\,C}\$ and \$Z_{\text{B}_i}=Z_{_\text{B}}=R\$. But it isn't necessary as we could reverse the roles just as easily or use distinct complex impedances for each. Doesn't really matter.
But for this purpose, we'll assume your situation, specifically, while still applying general rules to it.
I think you can see that \$V_0=V_1\cdot\frac{Z_{\text{B}_1}}{Z_{\large {\text{A}}_1}+Z_{\text{B}_1}}\$, since \$V_0\$ is the result of that specific voltage divider applied to \$V_1\$. This should not be hard to see.
Similarly, \$V_1=V_2\cdot\frac{Z_{\text{B}_2}\mid\mid\left(Z_{\large {\text{A}}_1}+Z_{\text{B}_1}\right)}{Z_{\large {\text{A}}_2}+Z_{\text{B}_2}\mid\mid\left(Z_{\large {\text{A}}_1}+Z_{\text{B}_1}\right)}\$ for similar reasons. The only complexity here is that you have to notice that this voltage divider has a slightly more complex expression since \$Z_{\text{B}_2}\$ is in parallel to \$Z_{ {\text{A}}_1}+Z_{\text{B}_1}\$.
And the same thing is true for \$V_1\$ except that now the voltage divider is still more complicated-looking because of the added components of the voltage divider here.
Suppose we say that each voltage divider can be expressed as \$\frac{Z_i}{Z_{\large {\text{A}}}+Z_i}=\frac1{ 1+\frac{Z_{\large {\text{A}}}}{Z_i} }\$, where \$Z_1=Z_{_\text{B}}\$, so that \$V_{_\text{OUT}}=V_{_\text{IN}}\cdot\prod_{i=1}^n \frac{Z_i}{Z_{\large {\text{A}}}+Z_i}\$. This can be turned into the following recurrences:
$$\begin{align*}
H_0&=1&&&G_0&=H_0
\\\\
H_i&=\frac1{2+\frac{Z_{\large {\text{A}}}}{Z_{\large {\text{B}}}}-H_{i-1}}&&&G_i&=H_i\cdot G_{i-1}
\end{align*}$$
In Python/Sympy:
def H(n,u):
if n < 1 : return 1
return 1/(2+u-H(n-1,u)
def G(n,u):
if n < 1 : return H(n,u)
return H(n,u)*G(n-1,u)
zc = 1/(s*C)
simplify(ratsimp(G(3,zc/R)))
C**3*R**3*s**3/(C**3*R**3*s**3 + 6*C**2*R**2*s**2 + 5*C*R*s + 1)
You can see that the numerator will be imaginary after substituting \$s=j\,\omega\$. So this means you want to select out the even powers of \$s\$ in the denominator, which will be the imaginary portion there taking into account the numerator, and set it to zero. While we are at it, let's also set \$\tau=R\cdot C\$.
Taking \$s=j\,\omega\$, collecting and then setting the even power terms equal to zero yields \$-6\,\tau^2\,\omega^2 + 1=0\$, or:
$$\begin{align*}6\tau^2\omega^2&=1
\\\\\tau^2\omega^2&=\frac16\\\\\tau\omega&=\frac1{\sqrt{6}}\\\\
\omega&=\frac1{\tau\sqrt{6}}\end{align*}$$
If you now substitute that into the above equation you will get a magnitude of exactly \$\frac1{29}\$:
abs(ratsimp(G(3,zc/R)).subs({s:I/R/C/sqrt(6)}))
1/29
Suppose this, instead, was an R+C rather than a C+R arrangement in the ladder. Then a similar process occurs. In this case, pick out the odd powers of \$s\$ since the numerator is real and not imaginary (after substitution.)
simplify(ratsimp(G(3,R/zc)))
1/(C**3*R**3*s**3 + 5*C**2*R**2*s**2 + 6*C*R*s + 1)
Taking \$s=j\,\omega\$, collecting and then setting the odd power terms equal to zero yields \$-\tau^3\,\omega^3 + 6\,\tau\,\omega=0\$, or \$\omega=\frac{\sqrt{6}}{\tau}\$ and we see confirmation of the same result using a different-valued \$\omega\$, as:
abs(ratsimp(G(3,R/zc)).subs({s:I/R/C*sqrt(6)}))
1/29
Returning back to your case, and assuming \$C=5.41\:\text{nF}\$ and \$R=1\:\text{k}\Omega\$, then \$\tau=5.41\:\mu\text{s}\$ and \$f=\frac1{2\pi\,5.41\:\mu\text{s}\,\sqrt{6}}\approx 12.01\:\text{kHz}\$.
Let's now ask the following questions:
How does the frequency change per Ohm of parasitic ESR for the capacitors? Since the numerator is already known to be imaginary, we want the real part of the denominator:
rpart = re(fraction(simplify(ratsimp(G(3,(ZC+re)/ZR))))[1].subs({s:I*omega}))
f0 = solve(Eq(rpart,0),omega)[0]/2/pi
diff(f0,re).subs({R:1e3,C:5.41e-9,re:1}).n()
-9.98944445144225
So we can see that a parasitic resistance of \$1\:\Omega\$ will lower the frequency by about \$10\:\text{Hz}\$.
How does the frequency change per %-change in C (using tolerance values?)
(diff(f0,C)*C/f0).subs({R:1e3,C:5.41e-9,re:0}).n()
-1.00000000000000
This means a +1% change in C will mean a -1% change in frequency.
How does the frequency change per %-change in R (using tolerance values?)
(diff(f0,R)*R/f0).subs({R:1e3,C:5.41e-9,re:0}).n()
-1.00000000000000
This means a +1% change in R will also mean a -1% change in frequency.
The last two might seem to be obvious and we probably should have expected it. But it's worth doing the math to make sure, too.
The first one might be interesting. Or not. But at least we have a quantitative clue.
