The answer to this question as asked is no. However, you generally obtain something similar.

Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $\sigma$ is the principal symbol of $D$, $\nu$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $D\Psi$ is orthogonal to $\ker(D)$ if and only if $$ \sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $\mathscr{R}$ is the space of $\Psi$ satisfying the orthogonality requirement $(*)$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where $$ \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned}$$ Here $\rho$, $\rho^*$ are the principal symbols of $Q$ and $Q^*$, respectively, so that $\sigma = \rho + \rho^*$.

In your specific example, it so happens that $L^k = M^k = \Gamma(X, \Lambda^k T^*X)$, while $\rho(\nu)$ and $\rho^*(\nu)$ are wedging with, respectively insertion of the normal vector. Therefore $\mathscr{T}$ and $\mathscr{N}$ have the description you gave.

As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $Q= D^+$ is (half of) the Dirac operator. In this case, $Q$ and $Q^* = D^-$ are elliptic, and $L^-$, $M^+$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $\pm$ instead of $0$, $1$ for the grading.)

The answer to this question as asked is no. However, you generally obtain something similar.

Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ (Of course $D\Phi = 0$ means that $D^2\Phi=0$. But $D^2\Phi = 0$ implies $$0 = \langle \Phi, (Q^*Q+QQ^*) \Phi \rangle = \|Q\Phi\|^2 + \|Q^*\Phi\|^2,$$ hence $Q\Phi = Q^*\Phi = 0$. In other words, $$\ker(D) \subseteq \ker (D^2) \subseteq \ker(Q)\cap \ker(Q^*).$$ The other direction is trivial.)

Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $\sigma$ is the principal symbol of $D$, $\nu$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $D\Psi$ is orthogonal to $\ker(D)$ if and only if $$ \sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $\mathscr{R}$ is the space of $\Psi$ satisfying the orthogonality requirement $(*)$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where $$ \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned}$$ Here $\rho$, $\rho^*$ are the principal symbols of $Q$ and $Q^*$, respectively, so that $\sigma = \rho + \rho^*$.

In your specific example, it so happens that $L^k = M^k = \Gamma(X, \Lambda^k T^*X)$, while $\rho(\nu)$ and $\rho^*(\nu)$ are wedging with, respectively insertion of the normal vector. Therefore $\mathscr{T}$ and $\mathscr{N}$ have the description you gave.

As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $Q= D^+$ is (half of) the Dirac operator. In this case, $Q$ and $Q^* = D^-$ are elliptic, and $L^-$, $M^+$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $\pm$ instead of $0$, $1$ for the grading.)

I believe that the answer to this question as asked is no. However, you generally obtain something similar.

Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $\sigma$ is the principal symbol of $D$, $\nu$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $D\Psi$ is orthogonal to $\ker(D)$ if and only if $$ \sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $\mathscr{R}$ is the space of $\Psi$ satisfying the orthogonality requirement $(*)$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where $$ \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned}$$ Here $\rho$, $\rho^*$ are the principal symbols of $Q$ and $Q^*$, respectively, so that $\sigma = \rho + \rho^*$.

In your specific example, it so happens that $L^k = M^k = \Gamma(X, \Lambda^k T^*X)$, while $\rho(\nu)$ and $\rho^*(\nu)$ are wedging with, respectively insertion of the normal vector. Therefore $\mathscr{T}$ and $\mathscr{N}$ have the description you gave.

As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $Q= D^+$ is (half of) the Dirac operator. In this case, $Q$ and $Q^* = D^-$ are elliptic, and $L^-$, $M^+$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $\pm$ instead of $0$, $1$ for the grading.)

The answer to this question as asked is no. However, you generally obtain something similar.

Consider $D = Q + Q^*$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $\sigma$ is the principal symbol of $D$, $\nu$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $D\Psi$ is orthogonal to $\ker(D)$ if and only if $$ \sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $\mathscr{R}$ is the space of $\Psi$ satisfying the orthogonality requirement $(*)$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where $$ \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned}$$ Here $\rho$, $\rho^*$ are the principal symbols of $Q$ and $Q^*$, respectively, so that $\sigma = \rho + \rho^*$.

In your specific example, it so happens that $L^k = M^k = \Gamma(X, \Lambda^k T^*X)$, while $\rho(\nu)$ and $\rho^*(\nu)$ are wedging with, respectively insertion of the normal vector. Therefore $\mathscr{T}$ and $\mathscr{N}$ have the description you gave.

As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $Q= D^+$ is (half of) the Dirac operator. In this case, $Q$ and $Q^* = D^-$ are elliptic, and $L^-$, $M^+$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $\pm$ instead of $0$, $1$ for the grading.)