(Since this question has been asked and highly upvoted multiple times over different posts, with no definitive answer as of yet, I thought it worthwhile to give a full, detailed answer.)
As shown in the question, the ideal $J$ generated by (non-leading) coefficients of the (generic) characteristic polynomial defines an irreducible variety, namely the set of nilpotent matrices. Since this variety has dimension $n^2 - n$ and $J$ has $n$ generators, $J$ is a complete intersection.
I will address the question of radicality, for complete intersections in general.
$\DeclareMathOperator{\n}{\mathfrak{n}} \DeclareMathOperator{\m}{\mathfrak{m}} \DeclareMathOperator{\codim}{codim}$
For a Noetherian ring, there are $2$ aspects to being reduced. The first is being generically reduced ($\iff$ Serre's condition $R_0 =$ regular in codimension zero), and the second is having no embedded primes ($\iff$ Serre's condition $S_1 =$ Cohen-Macaulay in codimension one).
For complete intersections, the Cohen-Macaulay condition comes for free. So one just needs to check regularity in codimension zero. Usually for finitely generated $k$-algebras this is done via the Jacobian criterion, but there is another approach, which I think deserves to be more well known.
What does it mean for a finitely generated $k$-algebra $R = k[X_i]/I$ to be generically reduced? In geometric terms (for $k$ algebraically closed): the Zariski tangent space at a general point $p \in V(I)$ should have dimension equal to that of $V(I)$. Restated algebraically: if $\m_p$ is the maximal ideal of $R$ corresponding to $p$, then $\dim_k \m_p/\m_p^2 \le \dim R$ (the other inequality always holds). Now $\m_p = \n_p/I$ for some maximal ideal $\n_p$ in $k[X_i]$, and
$$\m_p/\m_p^2 = (\n_p/I)/(\n_p/I)^2 = (\n_p/I)/((\n_p^2 + I)/I)\cong \n_p/(\n_p^2 + I) \cong (\n_p/\n_p^2)/((\n_p^2 + I)/\n_p^2)$$
as $k = R/\m_p = k[X_i]/\n_p$-vector spaces. This proves the following:
$\textbf{Theorem}$. Let $S = k[X_i]$ be a polynomial ring over an algebraically closed field $k$, $I = (f_1, \ldots, f_c) \subseteq S$ an ideal, and $p \in V(I)$ a general point, with maximal ideal $\n_p \subseteq S$. Then $I$ is generically reduced if and only if $\dim_k (\n_p^2+I)/\n_p^2 \ge \codim I$. In particular, if $c = \codim I$ (i.e. $I$ is a complete intersection), then $I$ is radical if and only if $f_1, \ldots, f_c$ are $k$-linearly independent mod $\n_p^2$, i.e.
\begin{equation}
\sum_{i=1}^c a_i f_i \in \n_p^2, \; a_i \in k \implies a_1 = \ldots = a_c = 0.
\end{equation}
Next, to apply this to the ideal $J$ in question: consider strictly upper-triangular matrices of the form
\begin{bmatrix}
0 & \ast & 0 & \ldots & 0 \\
0 & 0 & \ast & \ldots & 0 \\
\vdots & & \ddots & & \vdots \\
0 & 0 & \ldots & 0 & \ast \\
0 & 0 & \ldots & 0 & 0 \\
\end{bmatrix}
with $\ast \in k^\times$ nonzero: these are regular nilpotent matrices, which - up to conjugation by $GL_n$ - are dense in the set of all nilpotent matrices. Thus it suffices to check the theorem at such points $p$. As it happens, this can be done in a very clean combinatorial way: for such a point $p$,
$$\n_p = (X_{i,i+1} - b_i \mid 1 \le i \le n-1) + (X_{i,j} \mid 1 \le i, j \le n, j \ne i+1) \subseteq k[X_{ij}]$$
for some $b_1, \ldots, b_{n-1} \in k^\times$. If $A = (X_{ij})$ is the generic $n \times n$ matrix, and $f_i \in J$ is the coefficient of $\lambda^i$ in the characteristic polynomial of $A$, then (up to sign) $f_i$ is the trace of $\wedge^{n-i} A$, which is the sum of all principal $(n-i)$-minors of $A$ (e.g. $f_0 = \det(A)$, $f_{n-1} = \operatorname{tr}(A)$). Now mod $\n_p^2$, $f_i$ is a linear combination of $X_{n-i,1}, \ldots, X_{n,i+1}$ (namely the variables on the subdiagonal of $A$ of length $i+1$), where each coefficient is an $(n-1-i)$-fold product of $b$'s, hence is nonzero. (To see this, consider the expansion of $f_0 = \det A$: the only term which survives mod $\n_p^2$ is $X_{n,1} \prod_{i=1}^{n-1} X_{i,i+1}$, which is congruent to $X_{n,1} \prod_{i=1}^{n-1} b_i$. Finally, a principal minor of $A$ is nonzero mod $\n_p^2$ if and only if the row/column indices form an interval, in which case the corresponding principal submatrix of $p$ is again regular nilpotent, and the same reasoning with $f_0$ applies to the submatrix.)
In particular, $f_i$ mod $\n_p^2$ are linear forms, with disjoint supports for distinct $i$. This implies that $f_0, \ldots, f_{n-1}$ are linearly independent mod $\n_p^2$, so by the theorem above, $J$ is radical.
Additional remarks:
(i) It's useful to view radicality/primality criteria in a Noetherian ring from the viewpoint of primary decomposition. If $I = Q_1 \cap \ldots \cap Q_s$ is a minimal primary decomposition, and $P_i := \sqrt{Q_i}$ are the associated primes of $I$, with $P_1, \ldots, P_r$ the minimal primes, then:
$V(I)$ is irreducible iff $r = 1$, i.e. $I$ has a unique minimal prime
$I$ is generically reduced iff $Q_i = P_i$ for all $1 \le i \le r$
$I$ has no embedded primes iff $r = s$
It follows immediately that $I$ is radical if and only if (2) and (3) hold, and prime if and only if (1)-(3) hold.
(ii) Throughout the discussion on generic reducedness: if $V(I)$ is not irreducible, one should take multiple general points, one in each irreducible component of $V(I)$.
(iii) Although the stated theorem relies on $k$ being algebraically closed, the result about primeness of $J$ holds over any field (since base change to the algebraic closure is faithfully flat, and $J$ is defined over $\mathbb{Z}$).
Best Answer
We first note that any block matrix $ M =\left[\begin{matrix} A & B \\C & D \end{matrix}\right],$ where $A$ and $D$ are square matrices and $A$ is invertible, can be factorised in the form $$M = \left[\begin{matrix} A & B \\ C & D \end{matrix}\right] = \left[\begin{matrix} A & 0 \\ C & 1 \end{matrix}\right] \left[\begin{matrix} 1 & A^{-1}B \\ 0 & D - CA^{-1}B \end{matrix}\right], $$ so that $\det M = \det A \cdot \det(D - CA^{-1}B)$.
If we now calculate formally in the function field $\mathbb{C}(z)$, the upper-left block of $z \cdot I_{nm} - T$ is just $A = z\cdot I_n$, which is invertible over $\mathbb{C}(z)$. Using the above notation, one calculates $$D - CA^{-1}B = \left[\begin{matrix} z \cdot I_n & & & -b_1 - \frac 1 z b_0\\ -I_n & \ddots & &\vdots\\ & \ddots &z\cdot I_n&-b_{m-2}\\ & &-I_n & z \cdot I_n - b_{m-1} \end{matrix}\right]. $$ Now inductively we know that $$\det(D - CA^{-1}B) = \det(I_n z^{m-1} - b_{m-1}z^{m-2} - \dotsb - b_1 - \frac 1 z b_0).$$ Thus \begin{align*}\det (z I - T) & = z^n \det(I_n z^{m-1} - b_{m-1}z^{m-2} - \dotsb - b_1 - \frac 1 z b_0) \\ & = \det(I_n z^m - b_{m-1}z^{m-1} - \dotsb - b_1 z - b_0).\end{align*}