It is extremely unusual for the restriction of a Hilbert modular form to the diagonal to be an elliptic modular eigenform. It happens occasionally in some small cases (by coincidence, essentially), but there is nothing systematic which forces it to occur.

For instance, when $[K : \mathbf{Q}] = 2$, it turns out that there is a formula for the Petersson product $\langle res(\mathbf{f}), g\rangle$, where $\mathbf{f}$ is a Hilbert eigenform of weight $(k_1, k_2)$ over $K$, $res(\mathbf{f})$ is its diagonal restriction, and $g$ is an elliptic eigenform of weight $k_1 + k_2$. By a theorem of Paul Garrett, the product $\langle res(\mathbf{f}), g \rangle$ is (up to various fudge factors) equal to the square root of the central value of a degree 8 L-function -- the Rankin convolution of $g$ with the Asai L-function of $\mathbf{f}$. It would be extremely unlikely for this L-function to vanish for all but one of the possible eigenforms $g$ of the relevant level, so one wouldn't expect $res(\mathbf{f})$ to be an eigenform.

It is extremely unusual for the restriction of a Hilbert modular form to the diagonal to be an elliptic modular eigenform. It happens occasionally in some small cases (by coincidence, essentially), but there is nothing systematic which forces it to occur.

For instance, when $[K : \mathbf{Q}] = 2$, it turns out that there is a formula for the Petersson product $\langle res(\mathbf{f}), g\rangle$, where $\mathbf{f}$ is a Hilbert eigenform of weight $(k_1, k_2)$ over $K$, $res(\mathbf{f})$ is its diagonal restriction, and $g$ is an elliptic eigenform of weight $k_1 + k_2$. By a theorem of Paul Garrett (edit: and Harris--Kudla), the product $\langle res(\mathbf{f}), g \rangle$ is (up to various fudge factors) equal to the square root of the central value of a degree 8 L-function -- the Rankin convolution of $g$ with the Asai L-function of $\mathbf{f}$. It would be extremely unlikely for this L-function to vanish for all but one of the possible eigenforms $g$ of the relevant level, so one wouldn't expect $res(\mathbf{f})$ to be an eigenform.

This never happens.

Let $f$ be a Hilbert eigenform over some totally real field $K$. Then we have Hecke eigenvalues $a(\mathfrak{m})$ for every ideal $\mathfrak{m}$ of $K$. There's a certain amount of choice involved in defining what "restriction to the diagonal" means, but it must mean some series that looks like $\sum n^t a( n\mathcal{O}_K ) q^n$ (for some integer $t$), at least up to finitely many bad Euler factors

Now, there's a unique semisimple l-adic Galois representation for which the trace of Frobenius at $p$ is $p^t a(p \mathcal{O}_K)$ for all good primes $p$; it's the Asai, or twisted tensor product, representation attached to $f$, and it has dimension $2^{[K : \mathbf{Q}]}$. So if $K \ne \mathbf{Q}$ and the restriction to the diagonal is an eigenform, then we must have a 2-dimensional representation and a representation of dimension $> 2$ having the same trace, which is clearly absurd.

It is extremely unusual for the restriction of a Hilbert modular form to the diagonal to be an elliptic modular eigenform. It happens occasionally in some small cases (by coincidence, essentially), but there is nothing systematic which forces it to occur.

For instance, when $[K : \mathbf{Q}] = 2$, it turns out that there is a formula for the Petersson product $\langle res(\mathbf{f}), g\rangle$, where $\mathbf{f}$ is a Hilbert eigenform of weight $(k_1, k_2)$ over $K$, $res(\mathbf{f})$ is its diagonal restriction, and $g$ is an elliptic eigenform of weight $k_1 + k_2$. By a theorem of Paul Garrett, the product $\langle res(\mathbf{f}), g \rangle$ is (up to various fudge factors) equal to the square root of the central value of a degree 8 L-function -- the Rankin convolution of $g$ with the Asai L-function of $\mathbf{f}$. It would be extremely unlikely for this L-function to vanish for all but one of the possible eigenforms $g$ of the relevant level, so one wouldn't expect $res(\mathbf{f})$ to be an eigenform.